Question

For each of the following orders as listed in Column-I pick the correct observation listed in Colomn-II.
`{:("Column-I","Column-II"),((A)"CgtN",(p)"More favourable (exothermic)electron affinity"),((B)"SegtBr",(q)"The higher first ionization energy "),((C )"MggtK",(r)"The larger size"),((D)"FgtCl",(s)"The higher electronegativity"),(,(t)"The higher number of valence electrons"):}`

Answer 1

Correct Answer - A-p,r ; B-r ; C-q,s,t ; D-q,s
(A)Nitrogen has less favourable electron affinity because of stable half filled electron configuration `ns^2np^3` Nitrogen has higher first ionisation energy , because of stable half filled electron configuration , `ns^2np^3` Across the period , atomic size decreases , electrons are added in the same valence shell but for addition of each successive element the nuclear charge increases by one unit positive charge (i.e. proton) Electronegativity of nitrogen (3.0) is greater than that of carbon (2.5) on account of small atomic size.
`C=ns^2 np^2, N=ns^2 np^3`
(B)`Delta_(eg)H, Se= - 198 KJmol^(-1)` , Br = -`325 kJ mol^(-1)`
`Delta_(IE1)`,Se=941 kJ `mol^(-1)` , Br=1142 `kJ mol^(-1)`
Covalent radius/pm, Se=117, Br = 114
Electronegativity, Se=2.48 , Br = 3.0
Se=[Noble gas] `ns^2 np^4` , Br=[Noble gas] `ns^2 np^5`
(C )`Delta_(eg)`H, K=-48 kJ `mol^(-1)`, `Mg~~O`(stable configuration `ns^2`)
`Delta_(IE1), K 419 kJ mol^(-1)`, Mn`=737 kJ mol^(-1)`
Metallic radius /pm K=227, Mg=160
Electronegativity , K=0.8 , Mg=1.2
K =[Noble gas ] `ns^1` , Mg=[Noble gas]`ns^2`
(D)`Delta_(IE1)`,F=1680 kJ `mol^(-1)` , Cl = `1256 kJ mol^(-1)`
Covalent radius / pm , F=64 , Cl=99
Electronegativity , F=4.0 , Cl=3.2
Have same number of valence electrons because both belong to same group i.e. halogen