00:01
In this problem, we have to find that which molecule has the highest bond order.
00:10
We know that bond order is equal to one half of the difference in electrons present in bonding and anti -bonding molecular orbitals.
01:13
So bond order is equal to nb minus n a divide by 2 where nb indicates number of electrons present in bonding orbital and n a indicates number of electrons present in n t bonding molecular orbital first we have to write electronic configuration for molecular orbital to find bond order for each molecule for n2 dactonic configuration is sigma 1 is 2, sigma star 1 is 2, sigma 2 s 2 sigma 2 star 2s 2, pi 2 px 2, pi 2 p y 2 and sigma 2 p z 2.
02:20
Bond order for n2 equals to n b minus n a divide by 2.
02:29
Here nb equals to 10 and n a equals to 4 so we get bond order equals to 3 for o2 there is 16 electrons so electronic configuration is sigma 1 s 2 sigma star 1 is 2 sigma 2 s 2 sigma 2 starr 2 s 2 sigma 2 s 2 sigma 2 p z 2 pi 2 px2, pi 2, pi 2, pi star 2 px1, pi star 2 pyy 1.
03:20
Bond order for o2 is equal to nb minus n a divide by 2.
03:27
Here nb equals to 10 and n a equals to 6...