Question

From the following, number of compounds with \(sp^3d\) hybridisation are: \[ XeF_4,\; ICl_4^-,\; ICl_2^-,\; XeF_5^-,\; SF_4,\; XeF_2,\; ClF_3,\; BrF_5,\; NH_4^+ \]

Answer 1

Correct Answer: 4

Concept:
Hybridisation is determined by the steric number (sum of bonded atoms and lone pairs on the central atom). \[ \text{Steric number} = \text{bond pairs} + \text{lone pairs} \] If steric number \(=5\), hybridisation is \(sp^3d\). Step 1: Determine hybridisation of each compound. \[ XeF_4 : sp^3d^2 \] \[ ICl_4^- : sp^3d^2 \] \[ ICl_2^- : sp^3d \] \[ XeF_5^- : sp^3d^3 \] \[ SF_4 : sp^3d \] \[ XeF_2 : sp^3d \] \[ ClF_3 : sp^3d \] \[ BrF_5 : sp^3d^2 \] \[ NH_4^+ : sp^3 \] Step 2: Count molecules with \(sp^3d\). \[ ICl_2^-,\; SF_4,\; XeF_2,\; ClF_3 \] Total = \(4\) \[ \boxed{4} \]