Question

Arrange in decreasing order of stability- (a)CN (b)NO+ (c)O2 (d)CN- with reason plz​

Answer 1

Explanation:

STABLITY IS PROPORTIONAL TO BOND ORDER

SO CN⁻=NO>CN>O₂

Answer 2

The decreasing order of stability is as follows:

CN-=NO+>CN>O2.

• The Stability of the bond is directly proportional to the bond order of the molecule.
• "The bond order shows the number of chemical bonds present between a pair of atoms"
• Bond order =1/2[Bonding - antibonding]
• To calculate bond order we require electronic configuration, the electronic configuration for the given molecules are as follows:
• (CN)- - (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(σ2pz)2π2p2x=π2p2y

The electrons in bonding orbitals = 10

The electrons in antibonding orbitals = 4

Thus,

• Bond order =1/2[Bonding - antibonding]

Bond order=(10−4)/2=3

Therefore, the bond order of cyanide is 3.

• (NO)+ - σ2s)2 (σ*2s)2 (σ2pz)2 (π2px)2 (π2py)2 (π*2px)2 (π*2py)1

The electrons in bonding orbitals  = 8

The electrons in non bonding orbitals =2

Therfore,

• Bond order =1/2[Bonding - antibonding]

Bond order= 1/2(8 - 2)

Bond order = 1/2(6)

Bond order = 3

So, NO+ has bond order 3.

• (O2)- σ 1s2, σ *1s2, σ 2s2,  σ *2s2, σ 2p2z, 2p2x π = 2p2y π , 2p1x π *=2p1y π *

The electrons in bonding orbitals =10

The electrons in nonbonding orbitals=6

Therefore,

• Bond order =1/2[Bonding-antibonding]

Bond order = 12[10−6]=12(4)=2

Thus, the bond order O2 is 2.

• (CN)- It has 13 electrons

MO configuration : (σ1s) 2 ,(σ ∗ 1s) 2 ,(σ2s) 2 ,(σ ∗ 2s) 2 ,(π2p) 4 ,(σ2p) 1

The electrons in bonding orbitals = 9

The electrons in nonbonding orbital  = 4

So,

Bond order =1/2[Bonding-antibonding]

Bond Order= 1/2 1(9−4)=2.5

Therefore bond order of (CN) is 2.5.

Hence the order of stability is CN-=NO+>CN>O2.

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