Question

how many of the following molecules have sp³d hybridization on the central atom? sicl₄, brf₅, asf₅, brf₃

Explanation:

Step1: Calculate hybridization for \(SiCl_4\)

The formula for hybridization is \(H=\frac{1}{2}(V + M - C+A)\), where \(V\) is the number of valence electrons of the central atom, \(M\) is the number of monovalent atoms, \(C\) is the charge on cation and \(A\) is the charge on anion. For \(SiCl_4\), \(V = 4\), \(M=4\), \(C = 0\), \(A=0\). So \(H=\frac{1}{2}(4 + 4)=4\), hybridization is \(sp^{3}\)

Step2: Calculate hybridization for \(BrF_5\)

For \(BrF_5\), \(V = 7\), \(M = 5\), \(C=0\), \(A = 0\). \(H=\frac{1}{2}(7 + 5)=6\), hybridization is \(sp^{3}d^{2}\)

Step3: Calculate hybridization for \(AsF_5\)

For \(AsF_5\), \(V=5\), \(M = 5\), \(C = 0\), \(A=0\). \(H=\frac{1}{2}(5+5)=5\), hybridization is \(sp^{3}d\)

Step4: Calculate hybridization for \(BrF_3\)

For \(BrF_3\), \(V = 7\), \(M=3\), \(C = 0\), \(A=0\). \(H=\frac{1}{2}(7 + 3)=5\), hybridization is \(sp^{3}d\)

Answer:

2