(A) N2>N (Ist ionization energy)
As N2=14 electrons
Electronic configuration = (σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2
Electronic configuration of N=1s22s22p3
As we know, removal of an electron from fully filled orbital needs more energy, so,
N2>N(Ist Ionization energy)
(B) O2>O (Ist Ionization energy)
O2=(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(σ2pz)2(σ∗2Pz)2
O=1s22s22p4
As in O2, electrons present in antibonding is to be removed which is loosely held as we know antibonding orbital is higher in energy than atomic orbital. Higher in energy means less stability, electrons can be removed easily
(C) H2>H (Ist ionization energy)
H2=σ1s2H=σ1s1
Removal of an electron from fully filled orbital is tough hence, H2>H (Ist ionization energy)
(D) N+2>N−2(Stability)
B.O. of N+2 is 2.5
B.O. of N−2 is 2.5
But N−2 has more antibonding electrons than N+2 Hence, N+2 is more stable than N−2
(E) H2>H (Ist (Electron Affinity)
H2=σ1s2H=σ1s1
As in H2 orbital is already fully filled, hence, it will be very tough to add an electron.
(F) Na2CO3>Li2CO3 (Thermal statulity)
Li+ has an anamalously small size and very high charge density. Therefore, more polarising power. More polarized anion will decompose easily.
Hence, out of all, only 4 statements are correct i.e., A,C,D,F.