Online Test — Hydrogen
18 Questions • 15 min • Chapter MCQ
15:00
Question 1 of 18
The electronic configuration of a hydrogen atom is:
1s1
1s2
2s1
1s12s1
Explanation: Hydrogen has one electron in the 1s orbital: 1s1.
Question 2 of 18
Hydrogen resembles halogens because it can:
lose its electron easily
form a diatomic molecule and gain an electron to give H−
act only as a metal
form basic oxides
Explanation: Like halogens it is diatomic (H2) and is one electron short of a noble gas, forming H−.
Question 3 of 18
The number of neutrons in tritium is:
0
1
2
3
Explanation: Tritium (3H) has mass number 3 and one proton, so it has 2 neutrons.
Question 4 of 18
Water gas is a mixture of:
CO and H2
CO2 and H2
CH4 and CO
N2 and H2
Explanation: C + H2O → CO + H2; the mixture of CO and H2 is water gas.
Question 5 of 18
In the laboratory dihydrogen is prepared by reacting zinc with:
concentrated HNO3
dilute H2SO4
concentrated H2SO4
molten NaCl
Explanation: Zn + dilute H2SO4 → ZnSO4 + H2.
Question 6 of 18
Which of the following is a covalent (molecular) hydride?
NaH
CaH2
NH3
TiH1.7
Explanation: NH3 is formed by a p-block non-metal as discrete molecules, a covalent hydride.
Question 7 of 18
On electrolysis of a molten ionic hydride, hydrogen is liberated at the anode because hydrogen is present as:
H+
H2
H−
neutral H
Explanation: Ionic hydrides contain H−, which migrates to the anode and is oxidised to H2.
Question 8 of 18
The H–O–H bond angle in a water molecule is about:
90°
104.5°
109.5°
120°
Explanation: Lone-pair repulsion compresses the angle in the bent water molecule to about 104.5°.
Question 9 of 18
Temporary hardness of water is due to:
CaCl2 and MgCl2
CaSO4 and MgSO4
Ca(HCO3)2 and Mg(HCO3)2
NaCl
Explanation: Temporary hardness is caused by soluble bicarbonates of calcium and magnesium.
Question 10 of 18
Clark's method removes temporary hardness by adding:
washing soda
slaked lime, Ca(OH)2
common salt
alum
Explanation: Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O.
Question 11 of 18
Permanent hardness of water can be removed by:
boiling
the zeolite / ion-exchange process
cooling
decantation
Explanation: Permanent hardness (chlorides/sulphates) needs ion exchange; boiling does not work.
Question 12 of 18
When exhausted, sodium zeolite is regenerated by treating it with:
dilute HCl
concentrated NaCl (brine)
lime water
distilled water
Explanation: CaZ + 2NaCl → Na2Z + CaCl2; brine restores the sodium form.
Question 13 of 18
Heavy water is chiefly used as a:
fuel
moderator in nuclear reactors
bleaching agent
drying agent
Explanation: D2O slows fast neutrons without absorbing them, acting as a moderator.
Question 14 of 18
The oxidation state of oxygen in hydrogen peroxide is:
0
+1
−1
−2
Explanation: Each oxygen in H2O2 is in the −1 state, intermediate between 0 and −2.
Question 15 of 18
The structure of hydrogen peroxide is:
planar
linear
non-planar open-book
square planar
Explanation: H2O2 is non-planar with a dihedral (open-book) shape.
Question 16 of 18
When acidified KMnO4 is decolourised by H2O2, the peroxide acts as a(n):
oxidising agent
reducing agent
acid
dehydrating agent
Explanation: It reduces Mn(+7) to Mn2+ while being oxidised to O2, so it is a reducing agent.
Question 17 of 18
Hydrogen peroxide turns black PbS to white PbSO4. Here it acts as a(n):
reducing agent
oxidising agent
catalyst
dehydrating agent
Explanation: It oxidises sulphide to sulphate, so it acts as an oxidising agent (used in restoring paintings).
Question 18 of 18
The major product of the complete combustion of dihydrogen used as a fuel is:
CO2
water
carbon
hydrogen peroxide
Explanation: 2H2 + O2 → 2H2O; only water is formed, making it a clean fuel.