Question

Consider the following species:
\(BrF_5, XeF_5^-, BF_4^-, ICl_4^-, XeF_4, SF_4, NH_4^+, ClF_3, XeF_2, ICl_2^-\)
Number of species having \(sp^3d\) hybridized central atom is ____.

Answer 1

Correct Answer: 4

Step 1: Understanding the Question:
We need to find the number of species where the central atom has a steric number (SN) of 5, which corresponds to \(sp^3d\) hybridization.
Step 2: Key Formula or Approach:
Steric Number (SN) = \(\frac{1}{2}\) [Valence electrons + monovalent atoms - charge on cation + charge on anion].
Step 3: Detailed Explanation:
- \(BrF_5\): \(SN = \frac{1}{2}[7+5] = 6 \implies sp^3d^2\)
- \(XeF_5^-\): \(SN = \frac{1}{2}[8+5+1] = 7 \implies sp^3d^3\)
- \(BF_4^-\): \(SN = \frac{1}{2}[3+4+1] = 4 \implies sp^3\)
- \(ICl_4^-\): \(SN = \frac{1}{2}[7+4+1] = 6 \implies sp^3d^2\)
- \(XeF_4\): \(SN = \frac{1}{2}[8+4] = 6 \implies sp^3d^2\)
- \(SF_4\): \(SN = \frac{1}{2}[6+4] = 5 \implies sp^3d\) (1)
- \(NH_4^+\): \(SN = \frac{1}{2}[5+4-1] = 4 \implies sp^3\)
- \(ClF_3\): \(SN = \frac{1}{2}[7+3] = 5 \implies sp^3d\) (2)
- \(XeF_2\): \(SN = \frac{1}{2}[8+2] = 5 \implies sp^3d\) (3)
- \(ICl_2^-\): \(SN = \frac{1}{2}[7+2+1] = 5 \implies sp^3d\) (4)
The species are \(SF_4, ClF_3, XeF_2,\) and \(ICl_2^-\). Total = 4.
Step 4: Final Answer:
The number of species is 4.