Ut III Xii 2012 Chemistry Answer Key Vsp

8/12/2019 Ut III Xii 2012 Chemistry Answer Key Vsp

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DELHI PUBLIC SCHOOL GHAZIABAD

UNIT TEST III : 2012-13

SUBJECT : CHEMISTRY SET B

CLASS : XII

ANSWER KEY

Q NO. ANSWER MARKS

Q. 1 H2produced in reaction prevents oxidation of FeCl2to FeCl3 1

Q. 2 Zn, Cd, &Hg are not transition element 1

Q. 3 [ Co(NH3)(ONO)]2+

1

Q. 4 (i) 4NaCl + 2H2SO4 + MnO2 2 Na2SO4 + MnCl2 + 2H2O + Cl2

(ii) Cu2+

+ 4NH3 [Cu(NH3)4]2+

1

1

Q. 5 (i) IBr2-

Total ep is 5 ( 3are lp & 2bp) shape is Linear

(ii) XeO3 Total 4 ep(3 bp & 1lp) pyramidal

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1

Q. 6 (i) Configuration of M3+

= d3so n = 3

Magnetic moment = ( n( n + 2))1/2

= 151/2

= 3.87 B.M.

(ii) Cr2+

is strong reducing agent than Fe2+

because Cr3+

(t2g3) more

Stable than Fe3+

(d5)

1/2 +1/2

1

Q.7 Ni = 3d84s

2

Oxidation state of Ni = +2

Ni2+ =

Cl- is a weak ligand

[ NiCl4]2-

Sp3 tetrahedral paramagnetic

1/2

1/2

Q. 8 (i) Student A has = [Co(NH3)5SO4]Cl pentaamminesulphatocobalt(III)

chloride

Student B has = [Co(NH3)5Cl]SO4 pentaamminechloridocobalt(III)

sulphate

(or other structure in which C.no. of Co = 6 )

(ii) Value is identification, recogination ,learning by experimantation

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Q. 9 (i) E0reduction has atomization, I.E. & hydration energy for Cu

atomization energy is law & hydration energy is high

(ii) because d0is stable

(iii) For Mn2+

(d5) is stable but Fe

3+(d

5) & Cr

3+(t2g

3) states are stable

So Mn3+

/Mn2+

couple has more positive E0

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Q.10 Lanthanoid contraction : decrease in atomic size due to weak shielding by

4f electrons

Consequence (a) atomic size of 5d & 4d same

(b) I.E. of 5d is more than 3d& 4d ( or any two)

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Q. 11 (i) He is insoluble in blood

(ii) Sulphur in vapour state exit as S2 like O2so has 2 unpaired electron inAnti-bonding orbital

(iii) O is more electronegative than S

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Q. 12 (a) (i) HClO, < HClO2, < HClO3


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DELHI PUBLIC SCHOOL GHAZIABAD

UNIT TEST III : 2012-13

SUBJECT : CHEMISTRY SET A

CLASS : XIIANSWER KEY

Q NO. ANSWER MARKS

Q. 1 PCl3hydrolyse to liberate HCl so fume or by equation 1

Q. 2 Sc does not show variable oxidation state ( only +3) 1

Q. 3 Fe4[Fe(CN)6]3 1

Q. 4 (I) P4 + 3NaOH + 3H2O PH3 + 3NaH2PO2

(II) 2NaN3 2 Na + 3N2

1

1

Q. 5 (i) BrO3-Total 4 ep ( 3 are bp & 1 lp,) shape is pyramidal

(ii) XeF6 Total 7 ep (6 are bp & 1 lp) shape is distorted octahedral

1

1

Q. 6 (i) Configuration of M2+

= d7

so n = 3

Magnetic moment = ( n( n + 2))1/2

= 151/2

= 3.8 B.M.

(ii) In first half of the first transition series d orbital becomes more &

more half filled so + 2 state is stable than pairing of electron starts

1/2 +1/2

1

Q.7 Ni = 3d84s2Oxidation state of Ni = +2

Ni2+

=

CN-is a strong ligand so Ni

2+configuration changes in complex :

[ Ni(CN)4]2-

dsp2 square planer & diamagnetic

1/2

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Q. 8 (i) Student A has = [Co(NH3)5SO4]Cl pentaamminesulphatocobalt(III)

chlorideStudent B has = [Co(NH3)5Cl]SO4 pentaamminechloridocobalt(III)

Sulphate

(or any other with Co-ordination number of Co = 6)

(ii) Value is identification, recogination ,learning by experimentation

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Q. 9 (i) This is due to weak shielding by 5f electron than 4f .

(ii) Cu+disproportionate in aq. Solution

2 Cu+ Cu + Cu

2+

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(iii) Lower oxides are ionic so basic , higher oxides are covalent so acidic

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Q.10 Preparation of K2Cr2O7cromite ore of Cr:

(i) Ore to sod Na2CrO4

4FeOCr2O3 + 8Na2CO3(OR NaOH) + 7O2 2Fe2O3+ 8Na2CrO4 +8CO2

(ii) Conversion Na2CrO4 to Na2Cr2O72 Na2CrO4 + H2SO4 Na2SO4 + Na2Cr2O7+ H2O

(iii) Conversion of Na2Cr2O7to K2Cr2O7Na2Cr2O7 + 2KCl 2NaCl + K2Cr2O7 (full marks for unbalenced)

K2Cr2O7 on increasing pH converts to Na2CrO4 because at certain pH

both are at equil.

2

1

Q. 11 (i) H2SO4 H+ + HSO4

- (Ka1) : HSO4

- H

+ +SO4

2- ( Ka2)

In Ka2 H+is to be removed from anion

(ii) N-N bond energy is less than P-P

(iii) Due to small size of Br 7 F atoms are sterically hindered.

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Q. 12 (a) O3is passed through aq. Solution of KI & librated I2is estimated byStandered solution of sodium thio sulphate & starch as indicater

Eq. 2KI + O3 + H2O 2KOH + I2 + O2

(b) Contact process: S + O2 SO2

2SO2 + O2 2SO3 + heat

Reaction is reversible& exothermic so good yield of SO3 obtained at

Low T & high pressure & presence of catalyst Tt = 720 K /V2O5 catalyst

SO3 + H2SO4 dil. H2S2O7H2S2O7 + H2O 2H2SO4

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Q. 13 (I) Due to strong metallic bond.

(ii) availability of vacant orbital/ high nuclear charge density

(iii) unpaired electron in d orbital, d-d transition of electron in two

different set of d orbital is possible ,energy is in visible rang

(iv) small non-metal atoms like C,H can occupies interstitial site.

Q. 14 (a) [Fe(H2O)6]3+

Fe3+ = d5& H2O is a weak ligand so it has 5unpaired

Electrons, [Fe(CN)6]3-

Fe3+ d5but CN

-is a strong ligand so

It has 1 unpaired electron

(b) [Cr(en)3]3

optical isomers (dig.)

1+1

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BONUS QUESTIONA = MnO2 B = K2MnO

4 C = KMnO4

MnO2+ 4KOH + O2 2K2MnO4 + 2H2O

MnO42-

MnO4- + e

1 mark for correct compound A,B&C 2 marks for equation

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Ut III Xii 2012 Chemistry Answer Key Vsp