Coordination Compounds — Page 3 | NEET Chemistry

Q21

Pick out the correct statement with respect to [Mn(CN) 6 ] 3

-

.

A It is sp 3 d 2 hybridised and tetrahedral.

B It is d 2 sp 3 hybridised and octahedral.

C It is dsp 2 hybridised and square planar.

D It is sp 3 d 2 hybridised and octahedral.

Correct Answer

Option B

Solution

25 Mn = [Ar] 3d 5 4s 2 Mn +3 = [Ar] 3d 4 CN – is a strong field ligand thus, it causes pairing of electrons in 3d-orbital.

As, coordination number of Mn = 6, so it will form an octahedral complex.

Q22

The correct increasing order of trans -effect of the following species is

A NH 3 > CN

-

> Br

-

> C 6 H 5

-

B CN

-

> C 6 H 5

-

> Br

-

> NH 3

C Br

-

> CN

-

> NH 3 > C 6 H 5

-

D CN

-

> Br

-

> C 6 H 5

-

> NH 3

Correct Answer

Option B

Solution

The intensity of the trans-effect as measured by the increase in rate of substitution of the trans ligand) follows the sequence : CN $-$ > C 6 H 5 $-$ > Br $-$ > NH 3

Q23

Jahn-Teller effect is not observed in high spin complexes of

A d 7

B d 8

C d 4

D d 9

Correct Answer

Option B

Solution

Jahn-Teller distortion is generally significant for asymmetrically occupied e g orbitals as they are directed towards the ligands and the energy gain is more.

On the other hand in unevenly occupied t 2g orbitals, the John-Teller distortion is very weak.

Since the t 2g orbitals does not point directly at the ligand and thus energy gain is less.

Q24

Which of the following has longest C

-

O bond length? (Free C

-

O bond length in CO is 1.128

\mathop A\limits^o

.)

A [Fe(CO) 4 ] 2

-

B [Mn(CO) 6 ] +

C Ni(CO) 4

D [Co(CO) 4 ]

-

Correct Answer

Option A

Solution

[Fe(CO) 4 ] 2 $-$ has the lowest C—O bond order means the longest bond length.

Since metal atom is carrying maximum –ve charge therefore it would show maximum synergic bonding as a resultant C—O bond length would be maximum.

Q25

Number of possible isomers for the complex [Co(en) 2 Cl 2 ]Cl will be (en = ethylenediamine)

A 1

B 3

C 4

D 2

Correct Answer

Option B

Solution

$\therefore$ Total possible isomers = 3

Q26

The sum of coordination number and oxidation number of the metal M in the complex [M(en) 2 (C 2 O 4 )]Cl (where en is ethylenediamine) is

A 6

B 7

C 8

D 9

Correct Answer

Option D

Solution

[M (en) 2 (C 2 O 4 )]Cl x + 0 – 2 – 2 – 1 = 0 x = + 2 + 1 x = + 3 We know that coordination number is defined as the total number of binding sites attached to the metal.

$\therefore$ Coordination number = 6 Sum of oxidation number and coordination number = 3 + 6 = 9

Q27

The hybridization involved in complex [Ni(CN) 4 ]

2-

is (At. No. Ni = 28)

A sp 3

B d 2 sp 2

C d 2 sp 3

D dsp 2

Correct Answer

Option D

Solution

Electronic configuration of Ni 2+ : 3d 8 4s 0 Pairing of electrons in d-orbital takes place due to the presence of strong field ligand (CN - ).

Q28

The name of complex ion, [Fe(CN) 6 ] 3

-

is

A hexacyanitoferrate (III) ion

B tricyanoferrate (III) ion

C hexacyanidoferrate (III) ion

D hexacyanoiron (III) ion.

Correct Answer

Option C

Solution

[Fe(CN) 6 ] 3– x – 6 = –3 x = +6 – 3 x = +3 Hexa cyanido ferrate (III) ion.

Q29

Which of these statements about [Co(CN) 6 ] 3

-

is true?

A [Co(CN) 6 ] 3

-

has four unpaired electrons and will be in a high-spin configuration.

B [Co(CN) 6 ] 3

-

has no unpaired electrons and will be in a high-spin configuration.

C [Co(CN) 6 ] 3

-

has no unpaired electrons and will be in a low-spin configuration.

D [Co(CN) 6 ] 3

-

has four unpaired electrons and will be in a low-spin configuration.

Correct Answer

Option C

Solution

[Co(CN) 6 ] 3 $-$ oxidation no. of Co = +3 Co +3 = [Ar] 3d 6 4s 0 CN – is a strong field ligand and as it approaches the metal ion the electrons must pair up.

$\therefore$ [Co(CN) 6] 3– has no unpaired electrons and will be in a low spin configuration.

Q30

Cobalt (III) chloride forms several octahedral complexes with ammonia. Which of the following will not give test for chloride ions with silver nitrate at 25 o C?

A CoCl 3

\cdot

5NH 3

B CoCl 3

\cdot

6NH 3

C CoCl 3

\cdot

3NH 3

D CoCl 3

\cdot

4NH 3

Correct Answer

Option C

Solution

As For octahedral complexes, coordination number is 6.

CoCl 3 .

3NH 3 or [Co(NH 3 ) 3 Cl 3 ] will not give test for chloride ions with silver nitrate due to absence of ionisable chloride atoms.