In which of the following the oxidation number of oxygen has been arranged in increasing order?
OF2 < KO2 < BaO2 < O3
BaO2 < KO2 < O3 < OF2
BaO2 < O3 < OF2 < KO2
KO2 < OF2 < O3 < BaO2
B.
BaO2 < KO2 < O3 < OF2
Let the oxidation number of oxygen in following compound is x.
In OF2
x+(-1) 2 = 0
x = + 2
In KO2
+ 1+( x 2) = 0
2x =-1
x =-1/2
In BaO2
+2+ (x × 2) = 0
2x = -2
x =-1
In O3, oxidation number of oxygen is zero because oxidation number of an element in free state or in any of its allotropic form is always zero. Thus, the increasing order of oxidation number is
BaO2 < KO2 < O3 < OF2
-1 -1/2 0 +2
The electronic configuration of the element with maximum electron affinity is
1s2, 2s2,2p3
1s2, 2s2,2p5
1s2, 2s2,2p6, 3s2, 3p5
1s2, 2s2,2p6, 3s2, 3p3
The correct decreasing order of first ionisation enthalpies of five elements of the second period is
Be > B > C > N > F
N > F > C > B > Be
F > N > C > Be > B
N > F > B > C > Be
In which one of the following pairs the radius of the second species is greater than that of the first?
Na, Mg
O2-, N3-
Li+, Be2+
Ba2+, Sr2+
'Hydride Gap' is referred to which region of the Periodic Table ?
Groups 3, 4 and 5
Groups 5, 6 and 7
Groups 4, 5 and 6
Groups 7, 8 and 9
The lattice enthalpy and hydration enthalpy of four compounds are given below
Compound | Lattice enthalpy (in kJ mol-1) | Hydration enthalpy (in kJ mol-1) |
P | +780 | -920 |
Q | +1012 | -812 |
R | +828 | -878 |
S | +632 | -600 |
The pair of compound which is soluble in water is
P and R
Q and R
P and Q
Rand S
The first () and second (ΔH2) ionisation enthalpies (in kJ mol-1) and the (ΔegH) electron gain enthalpy (in kJ mol-1) of the elements I, II, III, IV and V are given below
Element | ΔH1 | ΔH2 | ΔegH |
I | 520 | 7300 | -60 |
II | 419 | 3051 | -48 |
III | 1681 | 3374 | -328 |
IV | 2372 | 5251 | +48 |
The most reactive metal and the least reactive non-metal of these are respectively
I and IV
IV and II
II and IV
III and IV
The first ionisation energy of oxygen is less than that of nitrogen. Which of the following is the correct reason for this observation?
Lesser effective nuclear charge of oxygen than nitrogen.
Lesser atomic size of oxygen than nitrogen
Greater interelectron repulsion between two electrons in the same p-orbital counter balances the increase in effective nuclear charge on moving from nitrogen to oxygen.
Greater effective nuclear charge of oxygen than nitrogen.
Column I | Column II | ||
(A) | He | (i) | High electron affinity |
(B) | Cl | (ii) | Most electropositive element |
(C) | Ca | (iii) | Strongest reducing agent |
(D) | Li | (iv) | Highest ionisation energy |
A-iv, B-i, C-ii, D-iii
A-iii, B-i, C-ii, D-iv
A-iv, B-iii, C-ii, D-i
A-ii, B-iv, C-i, D-iii