Coordination Compounds — Page 4 | NEET Chemistry

Q31

Which of the following complexes is used to be as an anticancer agent?

A mer -[Co(NH 3 ) 3 Cl 3 ]

B cis -PtCl 2 (NH 3 ) 2 ]

C cis -K 2 [PtCl 2 Br 2

D Na 2 CoCl 4

Correct Answer

Option B

Solution

Among the options listed, Option B, cis -PtCl 2 (NH 3 ) 2 ], is the one used as an anticancer agent.

This compound is commonly known as Cisplatin.

It's a platinum-based drug used in chemotherapy and is effective in treating various types of cancers, including testicular, ovarian, and bladder cancer.

The effectiveness of Cisplatin in cancer treatment lies in its ability to form platinum-DNA adducts, which interfere with DNA repair mechanisms, leading to cell apoptosis (programmed cell death) in rapidly dividing cells, such as cancer cells.

Q32

Among the following complexes the one which shows zero crystal field stabilization energy (CFSE) is

A [Mn(H 2 O) 6 ] 3+

B [Fe(H 2 O) 6 ] 3+

C [Co(H 2 O) 6 ] 2+

D [Co(H 2 O) 6 ] 3+

Correct Answer

Option B

Solution

CFSE for octahedral complex = [– 0.4 (t 2g electrons) + 0.6 (e g electrons)]

\Delta

0

Q33

Nitrogen detection in an organic compound is carried out by Lassaigne's test. The blue colour formed corresponds to which of the following formulae?

A Fe 3 [Fe(CN) 6 ] 2

B Fe 4 [Fe(CN) 6 ] 3

C Fe 4 [Fe(CN) 6 ] 2

D Fe 3 [Fe(CN) 6 ] 3

Correct Answer

Option B

Solution

The prussian blue colour is of Fe 4 [Fe(CN) 6 ] 3 ferric ferrocyanide.

Q34

The anion of acetylacetone (acac) forms Co (acac) 3 chelate with Co 3+ . The rings of the chelate are

A five membered

B four membered

C six membered

D three membered.

Correct Answer

Option C

Solution

Chelating ligands having conjugated double bonds form more stable six membered rings.

Q35

Which among the following is a paramagnetic complex?

A [Co(NH 3 ) 6 ] 3+

B [Pt(en)Cl 2 ]

C [CoBr 4 ]

2-

D Mo(CO) 6

Correct Answer

Option C

Solution

Co 2+ : [Ar]3d 7 4s 0 , here, Br – is a weak field ligand so will not cause pairing of d-electrons in Co 2+ .

$\therefore$ [CoBr 4 ] 2– will exhibit paramagnetic behaviour due to unpaired electrons.

Q36

Crystal field splitting energy for high spin d 4 octahedral complex is

A

-

1.2

\Delta

o

B

-

0.6

\Delta

0

C

-

0.8

\Delta

o

D

-

1.6

\Delta

o

Correct Answer

Option B

Solution

High spin d 4 = t 2g 3 e g 1 CFSE for octahedral complex = [– 0.4 (t 2g electrons) + 0.6 (e g electrons)]

\Delta

0 = (– 0.4 $\times$ 3 + 0.6 $\times$ 1)

\Delta

0 = (-1.2 + 0.6)

\Delta

0 = -0.6

\Delta

0

Q37

Which is diamagnetic?

A [Co(F) 6 ] 3

-

B [Ni(CN) 4 ] 2

-

C [NiCl 4 ] 2

-

D [Fe(CN) 6 ] 3

-

Correct Answer

Option B

Solution

Ni +2 = 3d 8 CN – is a strong ligand and causes pairing of 3d electrons of Ni 2+ .

As no unpaired electrons presents so it is diamagnetic.

Q38

The correct IUPAC name for [CrF 2 (en) 2 ]Cl is

A chloro difluorido ethylene diaminechromium (III) chloride

B difluoridobis (ethylene diamine) chromium (III) chloride

C difluorobis-(ethylene diamine) chromium (III) chloride

D chloro difluoridobis (ethylene diamine) chromium (III)

Correct Answer

Option B

Solution

IUPAC name of [CrF 2 (en) 2 ]Cl is Difluoridobis(ethylenediamine) chromium (III) chloride.

Q39

An excess of AgNO 3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of AgCl precipitated would be

A 0.003

B 0.01

C 0.001

D 0.002

Correct Answer

Option C

Solution

AgNO 3 + [Cr(H 2 O) 4 Cl 2 ]Cl $\to$ AgCl Excess mole = MV = 0.01 $\times$

{{100} \over {1000}}

= 0.001

Q40

A magnetic moment at 1.73 BM will be shown by one among of the following

A TiCl 4

B [CoCl 6 ] 4

-

C [Cu(NH 3 ) 4 ] 2+

D [Ni(CN) 4 ]2

-

Correct Answer

Option C

Solution

For $\mu$ = 1.73 B.M. n will be : 1.73 =

\sqrt {n\left( {n + 2} \right)}

$\Rightarrow$ n = 1 In [Cu(NH 3 ) 4 ] 2+ .

Cu +2 = [Ar]3d 9 NH 3 = Strong field ligand Here, unpaired electron gets excited to higher energy level but it still remains unpaired.