Question

1. the first ionization energy of Ca is greater than the first ionization energy of K, but the second ionization of Canis lower than the second ionization energy of K. Explain.

2. Compare the hybridization of the sulfur atom in the molecules SF_2, SF_4, and SF_6

3. Assign formal charges to each atom in NH_4^+, SO_3^-2, and XeF_2

Answer 1

1)

in calcium there are two paired electrons in 3s^2 orbital it is stable configuration so more energy is needed to remove an electrons from Ca. but in K there is only one unpaired electron present in 3s orbital so less amount of energy is required.

but the second ionisation energy of K is more than Ca. the reason is that in K+ ion it achieved octet configuration got nobel gas Ar . so we cannot easily remove electron from K+ ion .

2) SF2 ------------------> Sp3 hybridization

     SF4 -------------------> Sp3d

     SF6 ------------------> sp3d2

hybridization = central atom valency + monovalnet atoms / 2

suppose in SF2

hybridization = 6 + 2 / 2 = 4 (sp3)

3)

NH4+

formal charge of N = +1

formal charge of all H = +1

SO3-2 .

formal charge of S = 0

formal charge of one O = 0

formal charge of two O's = -1

XeF2

formal charge of F = 0

formal charge of Xe = 0