Match List I with List II : .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-c3owborder-color:inherit;text-align:center;vertical-align:top .tg .tg-0pkyborder-color:inherit;text-align:left;vertical-align:top List I (Complexes) List II (Hybridisation) A. [Ni(CO)_4] I. sp^3 B. [Cu(NH_3)_4]^2+ II. dsp ^2 C. [Fe(NH_3)_6]^2+ III. sp^3d^2 D. [Fe(H_2 O)_6]^2+ IV. d^2 sp^3

Coordination Compounds PYQ — JEE Main 2023 · 30 Jan · Shift 2 · Q4 with Solution

Match List I with List II:

List I (Complexes)		List II (Hybridisation)
A.	$\left[\mathrm{Ni}(\mathrm{CO})_{4}\right]$	I.	$\mathrm{sp}^{3}$
B.	$\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right]^{2+}$	II.	dsp $^{2}$
C.	$\left[\mathrm{Fe}\left(\mathrm{NH}_{3}\right)_{6}\right]^{2+}$	III.	$\mathrm{sp}^{3}\mathrm{d}^{2}$
D.	$\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+}$	IV.	$\mathrm{d}^{2} \mathrm{sp}^{3}$

A
A-I, B-II, C-III, D-IV
B
A-II, B-I, C-III, D-IV
C
A-I, B-II, C-IV, D-III
D
A-II, B-I, C-IV, D-III

View written solutionFree

Correct answer: C

Step 1: Determine hybridisation of each complex

We match each complex using oxidation state, electronic configuration, ligand strength, geometry, and whether inner/outer orbital complex is formed.

A. $[\mathrm{Ni}(\mathrm{CO})_4]$

Oxidation state of Ni: $x+4(0)=0 \Rightarrow x=0$
Electronic configuration of $\mathrm{Ni}$ : $\mathrm{Ni}: [\mathrm{Ar}]3d^8 4s^2$ So $\mathrm{Ni}^0$ effectively becomes: $3d^{10}$
Coordination number $=4$ , and carbonyl complexes are tetrahedral here.
Tetrahedral geometry corresponds to: $sp^3$

So, $A \to I$

B. $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$

Oxidation state of Cu: $x+4(0)=+2 \Rightarrow x=+2$
Electronic configuration of $\mathrm{Cu}^{2+}$ : $\mathrm{Cu}: [\mathrm{Ar}]3d^{10}4s^1$ $\mathrm{Cu}^{2+}: [\mathrm{Ar}]3d^9$
Coordination number $=4$ .
For $d^9$ Cu(II), four-coordinate ammine complex is typically square planar in this level of treatment.
Square planar hybridisation: $dsp^2$

So, $B \to II$

C. $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}$

Oxidation state of Fe: $x+6(0)=+2 \Rightarrow x=+2$
Electronic configuration of $\mathrm{Fe}^{2+}$ : $\mathrm{Fe}: [\mathrm{Ar}]3d^6 4s^2$ $\mathrm{Fe}^{2+}: [\mathrm{Ar}]3d^6$
$\mathrm{NH}_3$ is stronger than $\mathrm{H_2O}$ and can cause pairing here in standard coordination-compound matching questions.
Hence this is an inner orbital octahedral complex.
Inner orbital octahedral hybridisation: $d^2sp^3$

So, $C \to IV$

D. $[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}$

Oxidation state of Fe: $x+6(0)=+2 \Rightarrow x=+2$
Again, $\mathrm{Fe}^{2+}$ is: $3d^6$
$\mathrm{H_2O}$ is a weak field ligand, so no pairing occurs.
Therefore it forms an outer orbital octahedral complex.
Outer orbital octahedral hybridisation: $sp^3d^2$

So, $D \to III$

Step 2: Final matching

Thus, $A-I, \quad B-II, \quad C-IV, \quad D-III$

Step 3: Compare with options

This corresponds to Option C.

Step 4: Compare with stored correct answer

Stored correct answer: C

Our derived answer: C

So they agree.

More from Coordination Compounds

Step 1: Determine hybridisation of each complex

We match each complex using oxidation state, electronic configuration, ligand strength, geometry, and whether inner/outer orbital complex is formed.

[\mathrm{Ni}(\mathrm{CO})_4]

Oxidation state of Ni:

x+4(0)=0 \Rightarrow x=0

Electronic configuration of

\mathrm{Ni}

\mathrm{Ni}: [\mathrm{Ar}]3d^8 4s^2

\mathrm{Ni}^0

effectively becomes:

3d^{10}

Coordination number

=4

, and carbonyl complexes are tetrahedral here.

Tetrahedral geometry corresponds to:

sp^3

So,

A \to I

[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}

Oxidation state of Cu:

x+4(0)=+2 \Rightarrow x=+2

Electronic configuration of

\mathrm{Cu}^{2+}

\mathrm{Cu}: [\mathrm{Ar}]3d^{10}4s^1

\mathrm{Cu}^{2+}: [\mathrm{Ar}]3d^9

Coordination number

=4

For

d^9

Cu(II), four-coordinate ammine complex is typically square planar in this level of treatment.

Square planar hybridisation:

dsp^2

So,

B \to II

[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}

Oxidation state of Fe:

x+6(0)=+2 \Rightarrow x=+2

Electronic configuration of

\mathrm{Fe}^{2+}

\mathrm{Fe}: [\mathrm{Ar}]3d^6 4s^2

\mathrm{Fe}^{2+}: [\mathrm{Ar}]3d^6

\mathrm{NH}_3

is stronger than

\mathrm{H_2O}

and can cause pairing here in standard coordination-compound matching questions.

Hence this is an inner orbital octahedral complex.

Inner orbital octahedral hybridisation:

d^2sp^3

So,

C \to IV

[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}

Oxidation state of Fe:

x+6(0)=+2 \Rightarrow x=+2

Again,

\mathrm{Fe}^{2+}

is:

3d^6

\mathrm{H_2O}

is a weak field ligand, so no pairing occurs.

Therefore it forms an outer orbital octahedral complex.

Outer orbital octahedral hybridisation:

sp^3d^2

So,

D \to III

Coordination Compounds question

Coordination Compounds question

Step 1: Determine hybridisation of each complex

A. $[\mathrm{Ni}(\mathrm{CO})_4]$

B. $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$

C. $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}$

D. $[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}$

Step 2: Final matching

Step 3: Compare with options

Step 4: Compare with stored correct answer

More from Coordination Compounds

Coordination Compounds question

Coordination Compounds question

Step 1: Determine hybridisation of each complex

A. $[\mathrm{Ni}(\mathrm{CO})_4]$

B. $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$

C. $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}$

D. $[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}$

Step 2: Final matching

Step 3: Compare with options

Step 4: Compare with stored correct answer

More from Coordination Compounds

Step 1: Determine hybridisation of each complex

A. [Ni(CO)4][\mathrm{Ni}(\mathrm{CO})_4][Ni(CO)4​]

B. [Cu(NH3)4]2+[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}[Cu(NH3​)4​]2+

C. [Fe(NH3)6]2+[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}[Fe(NH3​)6​]2+

D. [Fe(H2O)6]2+[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}[Fe(H2​O)6​]2+

Step 2: Final matching

Step 3: Compare with options

Step 4: Compare with stored correct answer

More from Coordination Compounds

Step 1: Determine hybridisation of each complex

A. [Ni(CO)4][\mathrm{Ni}(\mathrm{CO})_4][Ni(CO)4​]

B. [Cu(NH3)4]2+[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}[Cu(NH3​)4​]2+

C. [Fe(NH3)6]2+[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}[Fe(NH3​)6​]2+

D. [Fe(H2O)6]2+[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}[Fe(H2​O)6​]2+

Step 2: Final matching

Step 3: Compare with options

Step 4: Compare with stored correct answer

More from Coordination Compounds

A. $[\mathrm{Ni}(\mathrm{CO})_4]$

B. $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$

C. $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}$

D. $[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}$

A. $[\mathrm{Ni}(\mathrm{CO})_4]$

B. $[\mathrm{Cu}(\mathrm{NH}_3)_4]^{2+}$

C. $[\mathrm{Fe}(\mathrm{NH}_3)_6]^{2+}$

D. $[\mathrm{Fe}(\mathrm{H_2O})_6]^{2+}$