Periodic Properties Test - 15

Sodium carbonate doesn't decompose on heating as it is stable to heat.

No. of electrons in O²⁻ = 10

No. of electrons in B³⁺ = 8

No. of electrons in Li⁺ = 4

No. of electrons in F⁻ =  10

F⁻ and O2⁻ have highest no. of electrons, thus the effective nuclear charge {nuclear charge/No. of electrons​} for them is less than B and Li.
When effective nuclear charge increases, size of an ion decreases.
Amongst F⁻ and O²⁻, size of O²⁻ is more than F⁻ as size decreases on moving from left to right in a period.
Therefore, O2⁻ is largest in size.

Atomic radii are measured in terms of Angstrom (Å)

Mg < Na < K < Rb. The order is based on the trend in a period and a group. The atomic radius decreases across a period (from left to right).

The atomic radius increases down a group due to the addition of a new energy level. Mg lies on the right side of Na and so it is smaller in size than that of Na.

K lies below Na in the same group and so, the size of K is greater than that of Na. And similarly, Rb is in the same group as K, Na and lies below K, Na.

We use n+l rule for calculating energies

value of l varies:

s = 0

p = 1

d = 2

f = 3

so  ns have energy n + 0 = n

(n-1)d have energy n -1 + 1 = n 

np have energy n + 1 

(n-1)f have energy n -1 + 2 = n + 1

if two shells have same value then we consider energy in given order

 s>p>d>f

we get: ns<(n−1)d<np<(n−1)f

The first ionization energy is the energy needed to remove the outermost, or highest energy, electron from a neutral atom in the gas phase. It is more for nitrogen than oxygen because nitrogen is more stable due to its half-filled electronic configuration.

Li has lowest ionization potential because after losing an electron it will attain noble gas configuration.

Atomic size increases as we move from top to down in a group, therefore, the amount of energy required for ejection of an electron from atom decreases, i.e. ionisation energy decreases. Hence, the correct order of IE₁​ is
Li>Na>K>Cs.

Isoelectronic means the species must have the same electronic configuration. 

→ Only the set of K⁺,Ca²⁺,Sc³⁺,Cl⁻ represents isoelectronic species. 

K, Z=19, Number of electrons in K⁺=18
Ca, Z=20, Number of electrons in Ca²⁺=18
Sc, Z=21, Number of electrons in Sc³⁺=18
Cl, Z=17, Number of electrons in Cl⁻=18

According to the values electron affinity the  three halogens X, Y and Z are Cl₂F₂ Br₂