Match Column-I with Column-II.

	Column-I		Column-II
(a)	Copper	(i)	Non-metal
(b)	Fluorine	(ii)	Transition metal
(c)	Silicon	(iii)	Lanthanoid
(d)	Cerium	(iv)	Metalloid

 

	(a)	(b)	(c)	(d)
1.	(ii)	(iv)	(i)	(iii)
2.	(ii)	(i)	(iv)	(iii)
3.	(iv)	(iii)	(i)	(ii)
4.	(i)	(ii)	(iii)	(iv)

The formula for the oxide formed by Eka-aluminum is:

1. ${\mathrm{EO}}_4$

2. ${\mathrm{E}}_2{\mathrm{O}}_3$

3. ${\mathrm{E}}_3{\mathrm{O}}_2$

4. EO

Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?

1.  F < Cl < O < S

2.  S < O < Cl < F

3.  O < S < F < Cl

4.  Cl < F < S < O

The ions O^2–, F^–, Na⁺, Mg²⁺, and Al³⁺ are isoelectronic. Their ionic radii show: 

The period and group number of the element with Z =114 are:

 The trend of atomic radius in a period and a group is :

An element with higher negative electron gain enthalpy in the given pair is:

(i) O or F  (ii) F or Cl
1. O, Cl
2. F, F
3. O, F
4. F, Cl  

The basic difference between the  electron gain enthalpy (E_a) and electronegativity (E_N) is:

The statement that "the electronegativity of N on the Pauling scale is 3.0 in all the nitrogen compounds" is incorrect because:

The general outer electronic configuration of s, p, d , and f-block elements respectively would be :