Dr. Sangeeta Khanna Ph.D 1 · The correct order of decreasing second ionization energies of Li, Be, B: a. Li > B > Be b. B > Li > Be c. Li > Be > B d. Be > B > Li A 20. Which of the

Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2017\+1\Physical\Grand Test\GT-3\+1-Grand Test-III.doc

Dr. Sangeeta Khanna Ph.D


Test Date: 12.8.2017 (Saturday) TOPIC: STRUCTURE OF ATOM & PERIODIC PROPERTIES

READ THE INSTRUCTIONS CAREFULLY

1. The test is of 2 hour duration. 2. The maximum marks are 358. 3. This test consists of 85 questions.

Section – A (Single Type Answer) Negative Marking [-1]

This Section contains 45 multiple choice questions. Each question has four choices A), B), C) and

D) out of which ONLY ONE is correct. 45 × 4 = 180 Marks

1. Which of the following is energy releasing process?

a. X– x(g) + e– b. O– (g) + e– O2–

c. O(g) O+(g) + e– d. O(g) + e– O–(g).

D

Sol. When an electron is added, energy is released due to forces of attraction (Electron Affinity)

2. Which of the following is true about the element with atomic number 18?

a. It has a very low ionization potential b. It has a very high electron affinity c. Its molecules are monoatomic d. Its electronegativity is very high C

Sol. Element with atomic number 18 is Noble gas (Ar). 3. Which of the following belongs to d-block element?

a. [Rn] 6d27s2 b. [Xe] 4f1 5d1 6s2 c. [Xe]4f14 5d1 6s2 d. [Xe] 5d1 6s2 D

Sol. Last electron in [Xe]5d1 6s2 goes to d-orbitals of last but one shell. 4. According to IUPAC norms a newly discovered element has been named as Uun. The atomic number

of the element is

a. 111 b. 112 c. 109 d. 110

D Sol. Unn refers to ununilium i.e., 5. Which of the following statements is most appropriate about effective nuclear charge.

a. It depend on the shielding constant

b. Depend only on atomic number

c. Depend on charge on the nucleus & electrons

d. Depend on both the nuclear charge and shielding constant

D

Sol. Effective nuclear charge = nuclear charge – screening constant. The value of screening constant is

calculated by using Slater’s Rule.

6. The maximum tendency to form the gaseous unipositive ion is for the element with configuration

a. 1s22s22p63s2 b. 1s22s22p63s23p1 c. 1s22s22p63s23p2 d. (Xe) 4f145d106s26p1 D

Sol. (d) has lowest IE and thus maximum tendency to loose electron. Inert pair effect (T)

7. Which of the following metal requires radiation of highest frequency to cause emission of electrons?

a. Na b. Mg c. K d. Ca B



Sol. Because it has highest IE out of the four given elements. 8. Out of the given alkali metals, the one with smallest size in water

a. Rb b. Cs c. K d. Na B

Sol. Cs will have smallest size due to least hydration 9. Paulings electronegativity values for elements are useful in predicting

a. Polarity of bonds in molecules b. Position of elements in electromotive series c. Co-ordination number d. Dipole moment of various molecules

A 10. Which block of the periodic table contains the element with configuration 1s22s22p63s23p63d104s1 a. s-block b. p-block c. d-block d. f-block C Sol. It is copper, a d – block transition element. 11. The ionic radii of N3–, O2–, F–, Na+ follows the order

a. N3– > O2– > F– > Na+ b. N3– > Na+ > O2– > F– c. Na+ > O2– > N3– > F– d. O2– > F– > Na+ > N3– A

Sol. All are isoelectronic ions. Size of iso electronic ions Z

1 (nuclear charge).

12. Which of the following statements is correct?

a. X– ion is larger size than X atom b. X+ ion is larger in size than X atom c. X+ ions is larger in size than X– ion d. X+ and X– ions are equal in size A

13. Among the following groupings which represents the collection of isoelectronic species?

a. NO+, C 22

, O2, CO b. N2, C 22

, CO, NO c. CO, N2, CN–, C 22

d. NO, CN–, N2, O 2

C Sol. (c) contains all isoelectronic species all have 14 electrons 14. Which one of the following arrangements not truly represent the property indicated against it?

a. Br < Cl < F : Electronegativity b. Br < F < Cl : Electron affinity c. Br2 < Cl2 < F2 : Bond energy d. Br2 < Cl2 < F2 : Oxidising power C

15. Consider the following changes:

A A+ + e– ; E1 and A+ A2+ + e– : E2 The energy required to pull out the two electrons are E1 and E2 respectively. The correct relationship

between two energies would be:

a. E1 < E2 b. E1 = E2 c. E1 > E2 d. E1 E2 A Sol. IE2 > IE1 or E1 < E2 16. Which one of the following is an incorrect statement? a. The ionization potential of nitrogen is greater than that of oxygen b. The electron affinity of fluorine is greater than that of chlorine c. The ionization potential of beryllium is greater than that of boron d. The electronegativity of fluorine is greater than that of chlorine B Sol. The electron affinity of fluorine is lower than that of chlorine.



17. First ionisation potential for copper is higher than that for potassium. The second ionization potential of copper is:

a. equal to the 2nd ionization potential of potassium b. more than the 2nd ionization potential of potassium c. less than the 2nd ionization potential of potassium d. Cannot be predicted C Sol. Cu (Z = 29) : 1s2 2s2 2p6 3s2 3p6 4s1 3d10 K (Z = 19) : 1s2 2s2 2p6 3s2 3p6 4s1 Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10 K+ : 1s2 2s2 2p6 3s2 3p6 Clearly second ionization energy of K will be more than Cu because in case of K the second electron

is to be removed from fully filled 3rd shell. 18. The one which is most basic out of the following: a. CO2 b. SiO2 c. Na2O d. SO2 C Sol. Metal oxides are basic. Out of C, Si, Na and S, only metal is Na. As such Na2O is most basic out of the

three. 19. The correct order of decreasing second ionization energies of Li, Be, B: a. Li > B > Be b. B > Li > Be c. Li > Be > B d. Be > B > Li A

20. Which of the following element has greatest difference between the first and second ionization enthalpies?

a. Na b. Mg c. Si d. P A 21. The successive ionization enthalpies of a main group element are 940, 2080, 3090, 4140, 5030, 6870,

16000 and 19500 kJ mol–1. The group to which the element belong is: a. 14 b. 15 c. 16 d. 17 C Sol. As greatest difference occurs between IE6 and IE7 the element has 6 electrons in the outermost shell

i.e., it belongs to 16th Group of the periodic table. 22. Assuming that elements are formed to complete the seventh period, what would be the atomic number

of the alkaline earth metal in the 8th period?

a. 113 b. 120 c. 119 d. 106 B

Sol. e– configuration 118(Nobel gas) 8s2

23. Which of the following isoelectronic ion has the lowest ionization energy? a. K+ b. Cl– c. Ca2+ d. S2– D Sol. It is biggest have least effective nuclear charge 24. Which of the following is arranged in decreasing order of size?

a. Mg2+ > Al3+ > O2– b. O2– > Mg2+ > Al3+ c. Al3+ > Mg2+ > O2– d. Al3+ > O2– > Mg2+ B

25. Which of the following statements is incorrect?

a. The first ionization energy of sulphur is greater than that of chlorine b. The third ionization energy of phosphorus is greater than that of aluminium c. The first ionization energy of aluminium is approximately the same as that of gallium d. The second ionization energy of boron is greater than that of carbon

A



26. In which of the following arrangements, the order is not correct according to the property indicated against it:

a. Increasing size: Al3+ < Mg2+ < Na+ < F– b. Increasing I.E.1 : B < C < N < O c. Increasing E.A.1 : I < Br < F < Cl d. Increasing metallic radius: Li<Na<K<Rb B

27. The correct order of increasing ionic character is

a. BeCl2 < MgCl2 < CaCl2 < BaCl2 b. BeCl2 < MgCl2 < BaCl2 < CaCl2 c. BeCl2 < BaCl2 < MgCl2 < CaCl2 d. BaCl2 < CaCl2 < MgCl2 < BeCl2 A

28. Ionisation potential of “X–(g)” is numerically equal to the electron affinity of a. X(g) b. X–(g) c. X+(g) d. X2–(g) A

29. In halogen group which tendency works with increase in atomic number?

a. Ionic size decreases b. Tendency to lose electrons decreases c. Ionization energy decreases d. In MX (M = metal, X = halogen) covalent character decreases.

C 30. In general the first ionisation energy decreases in a regular way on descending the main groups. A

departure from this trend is observed in Group 13 as follows:

Element Ionisation energy in KJ mol–1

1st 2nd 3rd

B 801 2427 3659

Al 577 1816 2744

Ga 579 1979 2962 In 558 1820 2704

Gallium has higher ionization energy in comparison to Al because a. gallium is harder than aluminium b. gallium is preceded by ten transition elements, where 3d shell is being filled which make Ga larger

than it would be c. gallium is preceded by ten transition elements where 3d shell is being filled which makes Ga

smaller than it would be d. effective nuclear charge for Ga is much higher than Al C

Sol. Ineffective shielding causes decrease in size. 31. The mass fraction of hydrogen in a compound of Group 14 element is 0.125. the hydride of this

element has formula: [Mass No. of Si – 28, C = 12, Ge = 72.5, Sn = 119] a. CH4 b. SiH4 c. GeH4 d. SnH4 B Sol. Let formula of hydride is AH4. Let atomic mass of A = a

Mass fraction of hydrogen in hydride = 28 a 125.0a4

14

Element A is Si(S28) and formula of hydride is SiH4. 32. Ionization energy is not influenced by:

a. Size of atom b. Effective nuclear charge c. Electrons present in inner shell d. Melting point D

Sol. Ionization energy is not affected by m.pt as it is the property of isolated gaseous atom.



33. If the nitrogen atom had electronic configuration 1s7, it would have energy lower than that of the normal ground state configuration 1s22s22p3, because the electrons would be closer to the nucleus. Yet, 1s7 is not observed because it violates.

a. Heisenberg uncertainty principle b. Hund’s rule c. Pauli exclusion principle d. Bohr postulates of stationary orbits C

Sol. This violates Pauli’s exclusion principle as an orbital cannot have more than two electrons. 34. Which of the following statement(s) are correct?

1. The electronic configuration of Cr is [Ar]3d54s1 (atomic No. of Cr = 24) 2. The magnetic quantum number may have a negative value 3. In silver atom, 23 electrons have a spin of one type and 24 of the opposite type (Atomic No. of

Ag = 47) 4. The oxidation state of nitrogen in HN3 is -3. a. 1, 2, 3 b. 2, 3, 4 c. 3, 4 d. 1, 2, 4 A

Sol. In HN3, the oxidation state is 3

1. Thus (4) is wrong statement.

35. The energy of electron in a hydrogen like species depends only on the ….. quantum number, while in multi-electron atoms it depends upon the …. and …… quantum number respectively.

a. n, , n b. , n, c. , n, m d. , m, n

A

36. 2zd orbital has

a. a lobe along Z axis and a ring along X-Y plane b. a lobe along Z axis and a lobe along X-Y plane c. a lobe along Z axis and a ring along Y-Z plane d. a lobe and a ring along Z axis. A

37. The angular momentum produced due to spin of electron is given by

a.

2

h

2

1 b.

2

h

2

3 c.

2

h

2

1 d.

2

h3

B

Sol. Angular momentum due to spin of electron = 2

1s as

2

h)1s(s

angular momentum =

2

h1

2

1

2

1

=

2

h

2

3

2

h

2

3

2

1

38. For 1s orbital, the probability of finding the electrons is

a. maximum in the nucleus b. maximum along the axis c. zero at the nucleus d. same in all directions but maximum around nucleus D

39. The ratio of the number of radial nodes to the total number of nodes for 2s orbital is

a. 1 : 2 b. 1 : 0 c. 1 : 1 d. 2 : 1 C

Sol. S-orbitals have only Radial nodes



40. The incorrect statement is

a. The most probable radius at which the electron will be found in H-atom is 53 pm

b. All orbitals with > 0 have zero probability of finding the electron at surface of nucleus

c. 2pz atomic orbital has xy as a nodal plane d. 2s atomic orbital has one angular node. D

41. For a 2s orbital for H-atom, the wave function is given by an equation

= 0a2/r

0

21

30

ea

r2

a8

1

where a0 = Bohr radius = 0.53 Å What is the value of the r in terms of a0 at the nodal point?

a. r = a0 b. r = 2a0 c. r = 2

a0 d. Zero

B

Sol. Nodal point means = 0, it is only if 0a

r2

0

i.e. r = 2a0

42. Mark out the incorrect options(s). Metal Characteristic colours in oxidizing flame a. Cesium Blue b. Calcium Brick Red c. Lithium Crimson Red d. Potassium Yellow D 43. The ratio of angular momentum of electron in two successive orbit is a (a > 1) and their difference is b.

Then a/b is equal to :

a. 1n

n

b.

n

1n c.

2

h.

n

1n d.

h

2.

n

1n

D

Sol. Angular momentum in two successive orbitals are 2

nh and

2

h1n

2

hb and

n

1na

h

2.

n

1n

b

a

44. Ionisation potential of hydrogen atom is 13.6 eV. If ground state of H-atom is excited by monochromatic radiations of 12.1 eV, then number of spectral lines emitted by H-atom on deexcitation will be:

a. 1 b. 2 c. 3 d. 4 C

Sol. 2n

n

6.13E Also E = En – E0

;

E = 13.6

22

2 n

1

n

1



12.1 = 13.6

22

2 n

1

1

1

12.1 = 13.6 - 22n

1

12.1 = 6.13n

6.13

2

n = 3

Thus, deexcitation will lead spectral lines = n = (3 – 1) = 3 45. A proton is accelerated to one tenth of the velocity of light. If its velocity can be measured with a

precision of 1%, then the uncertainty in its position is equal to or greater than a. 1.93 nm b. 19.3 nm c. 19.3 pm d. 193 pm D

Sol. m 1093.1(3.14) (4) )}s m 103)(kg101.9{(

s J0g626.6

4

h

p

1x 10

1-531

34

SECTION – B (Assertion and Reason) Negative Marking [– 1] This Section contains 10 questions. Each question has four choices A), B), C) and D) out of which

ONLY ONE is correct. 10 × 4 = 40 Marks

(a) If both assertion and reason are true and reason is the correct explanation of assertion. (b) If both assertion and reason are true but reason is not the correct explanation of assertion. (c) If assertion is true but reason is false. (d) If both assertion and reason are false.

1. Assertion :- Shielding effect increases as we go down the group. Reason : More is the number of electrons in the inner shell, more is the shielding effect. a. (a) b. (b) c. (c) d. (d) A 2. Assertion :- According to Fajan’s rule, covalent character is favoured by small cation and small anion.. Reason: The magnitude of covalent character in the ionic bond does not depends upon the extent of

polarization. a. (a) b. (b) c. (c) d. (d) D Sol. It is favoured by small cation and bigger anion. 3. Assertion: The first ionization enthalpy of aluminium is lower than that of magnesium. Reason: Ionic radius of aluminium cation is smaller than that of magnesium cation a. (a) b. (b) c. (c) d. (d) B 4. Assertion: The magnetic moment of Mg-atom is more than K-atom as the former has two electrons in

outermost shell. Reason: The magnetic moment of O-atom is more than magnetic moment of N-atom and former has

more number of unpaired electrons. a. (a) b. (b) c. (c) d. (d) D

Sol. Magnetic moment depends upon number of unpaired electrons and given by = )2n(n , where n is

no. of unpaired electrons.



Valence electron Unpaired electron K 1 1 Mg 2 0 N 3 3 O 4 2 5. Assertion: An electron in s-orbital is not circulating through the nucleus but simply distributed around

it whereas an electron in p-orbital can be thought of as circulating through the nucleus. Reason: For s-orbital angular momentum is zero for a p-orbital angular momentum is non-zero.

a. (a) b. (b) c. (c) d. (d) B

Sol. For S-orbital = 0; angular momentum =

2

h)1(

6. Assertion: d-orbitals are five fold non-degenerate in presence of magnetic field. Reason: In presence of magnetic field the energy of orbitals becomes altogether different. a. (a) b. (b) c. (c) d. (d) A Sol. Presence of magnetic field shows different repulsions for micromagnetic character developed in d-

orbitals due to different orientations.

7. Assertion: The 640 represents an orbital. Reason: The orbital will be 6 g. a. (a) b. (b) c. (c) d. (d) B

Sol. represents an orbital 640 means n = 6, = 4, m = 0, i.e., 6 g orbital

8. Assertion: In electronic configuration in Cr-atom, the 19th electron enters into 4s-atomic orbital but does not enter into 3d-atomic orbitals

Reason: Filling of electron follows “n + ” rule as per Aufbau’s principle. Electrons enter in that orbital

in which n + value is minimum

a. (A) b. (B) c. (C) d. (D) A 9. Assertion: The ground state of configuration of Cr is 3d54s1. Reason: A set of half filled orbitals containing one electrons each with their spin parallel provides extra

stability to the system. a. (A) b. (B) c. (C) d. (D) A Sol. Follow Hund’s rule

10. Assertion: The orbital angular momentum of d-orbital is 2

h6 .

Reason: Orbital angular momentum of electron is mvr = n2

h

a. (A) b. (B) c. (C) d. (D) C

Sol. Orbital angular momentum =

2

h)1( & for d orbital = 2

Orbital angular momentum = 2

h6

Orbit angular momentum of electron (mvr) = 2

nh



SECTION – C (Paragraph Type) Negative Marking [-1]

This Section contains 10 questions. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 10 × 4 = 40 Marks

Passage – 1 The diagram below shows part of the skeleton of the periodic table in which element are indicated by

letter which are not their usual symbols

Q

H L R

J T

M Z

Answer the following on the basis of periodic table : 1. Element having greatest ionic character in its compound a. T b. R c. M d. H C 2. Metal cation which is coloured in its aqueous solution a. H b. L c. J d. Q C 3. Element(s) of which carbonate salt is/are water soluble a. H b. L c. J d. M & H D 4. Which element is monoatomic gas at room temperature a. L b. Q c. R d. T D 5. Which of the element will not show stable +1 oxidation state a. H b. M c. Z d. Q D Passage – 2 Nuclear charge actually experienced by an electron is termed as effective nuclear charge. The

effective nuclear charge Z* actually depends on type of shell and orbital in which electron is actually present. The relative extent to which the various orbitals penetrate the electron clouds of other orbitals is.

s > p > d > f (for the same value of n) The phenomenon in which penultimate shell electrons act as screen or shield in between nucleus and

valence shell electrons and thereby reducing nuclear charge is known as shielding effect. The penultimate shell electrons repel the valence shell electron to keep them loosely held with nucleus. It is thus evident that more is the shielding effect, lesser is the effective nuclear charge and lesser is the ionization energy.

6. Which of the following valence electron experience maximum effective nuclear charge? a. 4s1 b. 4p1 c. 3d1 d. 2p3 D Sol. Electrons closer to nucleus will experience higher effective nuclear change. 2p3 is closer to 4s1 as principal quantum number is concerned first.



7. Which of the following is not concerned to effective nuclear charge?

a. Higher ionization potential of carbon than boron b. Higher ionization potential of magnesium than aluminium c. Higher values of successive ionization energy d. Higher electronegativity of higher oxidation state B

Sol. Magnesium having higher ionization potential due to more stable electronic arrangement [Ne] 3s2 in comparison to aluminium [Ne]3s23p1

Paragraph- 3

The letters n, , m proposed by Bohr, Sommerfeld and Zeeman respectively for quantization of angular

momentum in classical physics were later on obtained as the results of solution of Schrodinger wave

equation based on quantum mechanics. The term n, , m were named as principal quantum number,

azimuthal quantum number and magnetic quantum number respectively. The fourth quantum number s was given the name spin quantum number on the basis of two spins of electrons. The first two quantum numbers also decides the nodes of an orbital.

8. The numerical value 4, 3, 0 denotes :

a. 3d-orbitals b. 4f-orbitals c. 2s-orbitals d. 4d-orbitals B

Sol. represent an orbital and 4, 3, 0, has n = 4, = 3, i.e. 4f-orbital

9. The angular momentum of 3p-orbitals in terms of

2

h is:

a. 2 b. 2 c. 2

d.

2

A

Sol. Angular momentum in an orbital =

2

h.)1(

22

h)1(

10. Which statement about energy level in H-atom is correct?

a. Only n and decides energy level b. Only ‘’ decides energy level

c. Only n decides energy level d. n, and m decides energy level

C

Sol. Subshells of a shell in H-atom possess same energy level, i.e., does not specify for the energy level

of an orbital in H-atom.



SECTION – D (More than One Answer Type) No Negative Marking

This Section contains 10 questions. Each question has four choices A), B), C) and D) out of which MORE THAN ONE ANSWER is correct. 10 × 5 = 50 Marks

1. Among the given options, in which one does the first species have smaller size than the next one?

a. F–, O2– b. O2–, Na+ c. Al3+, Mg2+ d. O2–, O

A, C

2. Which is the correct order for the stated property?

a. Ba > Sr > Mg : Atomic radius b. F > O > N : First ionisation energy

c. Cl > F > I : Electron negativity d. O > Se > Te : Electronegativity

A,D

Sol. Electronic configuration of O is 1s22s22p2. Electronic configuration of N is 1s22s22p3. As electron is to

be removed from half filled shell (2p3) in case of N. So, first ionization energy of N is more than that of

oxygen.

3. Consider the following ionization steps:

M(g) M+(g) + e–; H = 200 eV M(g) M2+(g) + 2e–; H = 400 ev

Select correct statement(s)

a. I.E.1 of M(g) is 200 eV b. I.E. of M+(g) is 200 eV

c. I.E.2 of M(g) 400 eV d. Sum of first two I.E. is 400 eV

A,B,D

4. Which of the following processes involves absorption of energy?

a. S(g) + e– S– (g) b. S–(g) + e–

S2–(g)

c. Cl(g) + e– Cl–(g) d. Ne(g) + e– Ne(g)

B,D

5. Which of the following is/are correctly matched?

a. F2 > Cl2 > Br2 > I2 Bond dissociation energy

b. Mn2O7 > CrO3 > V2O5 Acidic strength

c. Cl > F > Br > I Ionizaton energy

d. Al+3 > Mg+2 > Na+ Hydration energy

B,D

6. Which of the following is correct order of acidic strength.

(a) H2S < PH3 < HCl (b) H2SO3 < H2SO4 < HClO4

(c) HF < HCl < HBr < HI (d) H2S < H2Se < H2Te

B,C,D 7. Which of the following are correct?

a. Each atom has at least one orbital symmetrical about the nucleus b. Each orbit has at least one orbital symmetrical about the nucleus c. Number of electrons in Ne having their angular momentum equal to zero are four d. Number of waves made by an electron in an orbit is equal to number of orbit

A,B,C,D

Sol. (a) ‘S’ orbital is symmetrical & each orbit have S-orbital.

(c) Ne have 1s22s2 = 4e– with = 0

(d) No. of wave in an orbit = n



8. Which of the following statements are correct?

a. The probability of finding a 4d electron right at the nucleus is zero

b. For all values of m, the p-orbital have the same shape, but overall size increases as n increases,

for a given atom.

c. A 2px atomic orbital consists of two lobes of electron density

d. There is no probability of finding a p-electron right at the nucleus

A,B,C,D 9. Which of quantum numbers are consistent with the theory?

a. n = 2, = 1, m = 0, s = +1/2 b. n = 4, = 3, m = -2, s = -1/2

c. n = 3, = 2, m = -3, s = +1/2 d. n = 4, = 3, m = -3, s = +1/2

A, B, D

Sol. In ‘C’ m cannot have value more than

10. Choose the correct statement:

a. The shape of an atomic orbital depends upon azimuthal quantum number b. The orientation of an atomic orbital depends upon the magnetic quantum number c. The energy of an electron in an atomic orbital of multi-electron atom depends only on principal

quantum number d. The number of degenerate atomic orbitals of one type depends upon the value of azimuthal A,B,D

Sol. (d) No. of orbital in a sub level = 2 + 1

(c) energy of orbital depend on n & in multielectron system.

SECTION – E (Matrix Type) No Negative Marking

This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. 2 × 8 = 16 Marks

1. Match Column – I with Column – II

Column – I Column – II

(A) Flourine (P) Gas at room temperature

(B) Chlorine (Q) Highest electronegativity in periodic table

(C) Bromine (R) Highest electron affinity in periodic table

(D) Iodine (S) Can show positive oxidation state

Sol. A P, Q; B P, R, S; C S; D S

F2(g), Cl2(g), Br2(), I2(s)

Flourine show only 0 and – 1 oxidation state. 2. Match Column – I with Column – II [At. No. = Cr = 24; Zn = 30; Cu = 29; Pd = 46]

Column – I Column – II

(A) Cr+3 (P) Paramagnetic

(B) Zn+2 (Q) have d10 configuration

(C) Cu+1 (R) Diamagnetic

(D) Pd (S) have 3 unpaired electron

Sol. A P, S; B Q, R; C R, Q; D Q, R



SECTION – F (Integer Type) No Negative Marking

This Section contains 8 Questions. The answer to each question is a Single Digit Integer ranging from 0 to 10. The correct digit below the question number in the OMR is to be bubbled. 8 × 4 = 32

1. The first three successive ionization energies of an element Y are 900, 1757 and 14850 kJ mol–1

respectively. The element Y belongs to groups

Sol. 2

As E1 is not very high element Y cannot belong to group 15 or group 17. As successive IE3 increases in

the order

E1 < E2 < < < E3

Element Y belongs to group 2.

2. How many of following elements can show variable valency.

Na, Mg, Fe, Zn, P, S, Sn, Pb

Sol. 5 [Fe, P, S, Sn, Pb]

3. How many of following element will have fully filled orbitals with = 2. Fe, Zn, Cu, Ag, Cd, La, Pd, Hg

[At. No. Fe = 26, Cu = 29, Zn = 30, Ag = 47, Cd = 48 ; Pd = 46 ; La = 57, Hg = 80]

Sol. 6

= 2 is

Zn = 3d104s2; Cu = 3d104s1; Ag = 4d105s1; Cd = 4d105s2; Pd = 4d105s0; Hg = 5d106s2

4. An element has ‘8’ electrons each in both valence as well as penultimate shells. In this element total number of electrons which has zero value of orbital angular momentum is:

Sol. 6

Element should be argon (18)

Zero value of orbital angular momentum means ‘s’ electrons. 1s2, 2s22p6, 3s23p6 [2, 8, 8] 5. From the following elements the total number of elements which have ‘ns1’ configuration in their

valence shell: Na, N, Sc, Cr, Cu, Ag, Cs, Mo and Zn.]

(At. No. = Sc [21]; Cr [24]; Cu [29]; Ag [47]; Cs [55]; Mn [25]) Sol. 6

Na 1s2, 2s22p6, 3s1 Na and Cs are alkali metals

N 1s2, 2s22p3 Cu and Cr are exceptions of 3d series

Sc (Ar) 3d1 4s2

Zn (Ar) 3d10 4s2

6. An element has total 12 electrons in ‘p’ orbitals and total 8 electrons in ‘d’ orbitals. Total number of electrons in ‘s’ orbitals would be:

Sol. 8

12 electrons in ‘p’ orbitals = 2p6, 3p6 8 electrons in ‘d’ orbitals = 3d8 1s2, 2s22p6, 3s2 3p6, 4s2 3d8

Cr

Cu

(Ar) 3d5 4s1

(Ar) 3d10 4s1

In ‘4d’ series Nb, Mo, Tc, Ru, Rh, Ag have 5s configuration.

Ag

Cs

Mn

(Kr) 4d10 5s1

(Xe) 5s1

3d5 4s

2

In ‘5d’ series only Pt and Au have 6s1 configuration



7. Which of the following statement is not correct for Lithium

(a) Lithium has highest oxidation potential

(b) Lithium always form ionic compound

(c) Lithium form stable peroxide & superoxide

(d) LiNO3 on heating give Brown gas NO2

(e) Lithium carbonate is thermally stable

(f) Lithium does not show photoelectric effect

(g) Lithium give yellow colour in flame

(h) Lithium has lowest density in its group

(i) Lithium form both oxide and nitride when burn in air like Mg.

(j) Lithium ion has highest hydration energy in its gp

Sol. 4 (b, c, e, g)

(b) Li form ionic as well as covalent compound

(c) Li form stable oxide

(e) Li carbonate is thermally unstable

(g) Li give red colour in flame 8. An electron in the ground state of H atom absorbs energy equal to 10.2 eV. Find the excited state to

which the electron will be promoted. Sol. 1 Energy of electron in ground state = -13.6 eV Energy supplied = 10.2 eV T.E. = -3.4

En = -13.6 2

2

n

ZeV

-13.6 2

2

n

ZeV = -3.4 eV

2n

6.13eV = -3.4 eV

n2 = 4.3

6.13 ; n2 = 4 ; n = 2

Documents

Dr. Sangeeta Khanna Ph.D 1 · The correct order of decreasing second ionization energies of Li, Be, B: a. Li > B > Be b. B > Li > Be c. Li > Be > B d. Be > B > Li A 20. Which of the