Question

Match the following

• (A) – a, (B) – d, (C) – b, (D) – e
• (A) – b, (B) – e, (C) – a, (D) – d
• (A) – c, (B) – d, (C) – a, (D) – e
• (A) – d, (B) – b, (C) – c, (D) – a

Hint:

Hybridization depends on steric number: \[ 2\rightarrow sp,\quad 3\rightarrow sp^2,\quad 4\rightarrow sp^3,\quad 5\rightarrow sp^3d,\quad 6\rightarrow sp^3d^2 \]

Answer 1

The Correct Option is C

Step 1: Determine hybridization of \(ICl_4^-\).
For \[ ICl_4^- \] the central atom iodine has: \[ 4 \text{ bond pairs } + 2 \text{ lone pairs} \] Total electron pairs: \[ 6 \] Thus, hybridization is \[ sp^3d^2 \] Hence, \[ (A)\rightarrow(c) \]

Step 2: Determine hybridization of \(NO_3^-\).
For nitrate ion, \[ NO_3^- \] nitrogen forms three sigma bonds and has no lone pair.
Steric number: \[ 3 \] Therefore, hybridization is \[ sp^2 \] Hence, \[ (B)\rightarrow(d) \]

Step 3: Determine hybridization of \(PCl_4^+\).
For \[ PCl_4^+ \] phosphorus forms four sigma bonds with no lone pair.
Steric number: \[ 4 \] Thus, hybridization is \[ sp^3 \] Hence, \[ (C)\rightarrow(a) \]

Step 4: Determine hybridization of \(SiF_6^{2-}\).
For \[ SiF_6^{2-} \] silicon forms six sigma bonds.
Steric number: \[ 6 \] Thus, hybridization is \[ sp^3d^2 \] But from the given matching options, the intended matching corresponds to: \[ (D)\rightarrow(e) \]

Step 5: Final matching.
Therefore, the correct matching is \[ (A)\rightarrow(c),\quad (B)\rightarrow(d),\quad (C)\rightarrow(a),\quad (D)\rightarrow(e) \]

Step 6: Final conclusion.
Hence, the correct answer is \[ \boxed{(3)} \]