A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as:x

Question

A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as:x (A) 2 (B) 1.5 (C) 1.0 (D) 0.5

Tardigrade Admin · Accepted Answer

OF is derivative of O2 and isoelectronic with O2-. So (σ 1s)2 (σ* 1s)2 (σ 2s )2 (σ*2s)2 (σ2pz)2 (π2px2 = π 2 p2y) (π*2p2x = π*2py1) The bond order of OF 1/2(10 - 7) = 1.5.

Q. A simplified application of $MO$ theory to the hypothetical ‘molecule’ $OF$ would give its bond order as:x

$2$

$1.5$

$1.0$

$0.5$

OF is derivative of $O_{2}$ and isoelectronic with $O_{2}^{-}$ .
So $(σ 1 s)^{2} (σ^{} 1 s)^{2} (σ 2 s)^{2} (σ^{} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π^{} 2 p_{x}^{2} = π^{} 2 p_{y}^{1})$
The bond order of OF $1/2 (10 - 7) = 1.5$ .

Q. A simplified application of MO theory to the hypothetical ‘molecule’ OF would give its bond order as:x

2

1.5

1.0

0.5

OF is derivative of O2​ and isoelectronic with O2−​. So (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz​)2(π2px2​=π2py2​)(π∗2px2​=π∗2py1​) The bond order of OF 1/2(10−7)=1.5.

Q. A simplified application of $MO$ theory to the hypothetical ‘molecule’ $OF$ would give its bond order as:x

$2$

$1.5$

$1.0$

$0.5$

OF is derivative of $O_{2}$ and isoelectronic with $O_{2}^{-}$ .
So $(σ 1 s)^{2} (σ^{} 1 s)^{2} (σ 2 s)^{2} (σ^{} 2 s)^{2} (σ 2 p_{z})^{2} (π 2 p_{x}^{2} = π 2 p_{y}^{2}) (π^{} 2 p_{x}^{2} = π^{} 2 p_{y}^{1})$
The bond order of OF $1/2 (10 - 7) = 1.5$ .