Consider the following species, CN^(+), CN^(-1), NO and CN Which one of these will have the highest bond order?

Question

MARKS Content Team · Accepted Answer

Correct answer: (B) CN^(-). Ion/Species Total electron Bond order NO 15 2.5 CN^(-) 14 3 CN^(+) 12 2 CN 13 2.5 Number of e^(−) in NO=7+8=15 NO=σ1s^2<σ *1s^2<σ2s^2<σ *2s^2<σ2Pz2<π2Py2=π2Px^2<π*2Py1=π *2Px0 Bond order =((number of electron in bonding molecular orbital − number of electron in anti bonding molecular orbital))/2 BO=(6−1)/2=2.5 . Number of e^(−) in CN=6+7=13 NO=σ1s^2<σ *1s^2<σ2s^2<σ *2s^2<σ2Pz2<π2Py2=π2Px^2<π*2Py1=π *2Px0. Bond order =5/2 =2.5 Number of e^(−) in CN^(+)=6+7−1=12 NO=σ1s^2<σ *1s^2<σ2s^2<σ *2s^2<σ2Pz2<π2Py2=π2Px^2<π*2Py1=π *2Px0 Bond order =(4−0)/2 =2 CN^(−)= number of e^(−) in CN^(−)=6+7+1=14 CN^(-)=σ1s^2<σ *1s^2<σ2s^2<σ *2s^2<σ2Pz2<π2Py2=π2Px^2<π*2Py1=π *2Px0 Bond order=(6−0)/2=3 . CN^(−) as highest bond order.

Ion/Species	Total electron	Bond order
$NO$	$15$	$2.5$
${CN}^{-}$	$14$	$3$
${CN}^{+}$	$12$	$2$
$CN$	$13$	$2.5$

Consider the following species,

${CN}^{+}, {CN}^{- 1}, NO and CN$

Which one of these will have the highest bond order?

Solution

Relevant Questions

Consider the following species,CN+, CN-1, NO and CNWhich one of these will have the highest bond order?

Solution

Relevant Questions

Consider the following species,

${CN}^{+}, {CN}^{- 1}, NO and CN$

Which one of these will have the highest bond order?