Q.1. The element with atomic number 26 will be found in group :
(a) 2
(b) 8
(c) 6
(d) 10
Correct Answer is option (b)
The valence shell configuration of Z = 26 is [Ar] 3d6 4s2. That for the group of element is 6 + 2 = 8
Q.2. Mode of hybridization affects
(a) ‘s’ character of a bond pair electron
(b) ‘s’ & ‘p’ character of a bond pair
(c) None of these
(d) Odd electron
Correct Answer is option (b)
Hybridised orbital accommodates σ bond pairs.
Q.3. The elements with atomic numbers 9, 17, 35, 53, 85 are all ———-
(a) halogens
(b) noble gases
(c) alkali earth metals
(d) transition metals
Correct Answer is option (a)
The elements of atomic numbers 9, 17, 35, 53, 85 are respectively F, Cl, Br, I, At etc are called halogen.
Q.4. The correct order of the size of the iodine species is:
(a) I >I+> I-
(b) I > I--> I+
(c) I+> I–> I
(d) I–> I >I+
Correct Answer is option (a)
Size of an anion > size of an atom > size of a cation
Q.5. Which of the following transitions involves maximum amount of energy?
(a) M–(g) → M(g)
(b) M(g) → M+(g)
(c) M+(g) → M2+(g)
(d) M2+(g) → M3+(g)
Correct Answer is option (a)
Removal of an electron from a dipositive ion involves maximum amount of energy.
Q.6. Which of the following electronic configurations of an atom has the lowest ionisation enthalpy?
(a) 1s2 2s2 2p3
(b) 1s2 2s2 2p6 3s1
(c) 1s2 2s2 2p6
(d) 1s2 2s2 2p5
Correct Answer is option (b)
Ionisation enthalpy is the amount of energy required when an electron is removed from the outermost orbit of an isolated gaseous atom. Electronic configuration of 1s2 2s2 2p6 3s1 has lowest ionisation enthalpy.
Q.7. Which is least soluble in water
(a) CaSO4
(b) MgSO4
(c) Na2SO4
(d) BaSO4
Correct Answer is option (d)
Solvation energy > lattice energy.
Q.8. The Ionic radius of cation is always ________.
(a) Less than the atomic radius
(b) more than the atomic radius
(c) Equal to atomic radius
(d) Cannot be predicted
Correct Answer is option (a)
The Ionic radius of cation is always less than the atomic radius. Cation is formed by the loss of electrons. So that the effective nuclear charge increases as a result ionic radius decreases.
Q.9. Among NaF, NaCl, NaBr and Nal, the NaF has highest melting point because
(a) It has maximum ionic character
(b) It has minimum ionic character
(c) It has associated molecules
(d) It has least molecular weight
Correct Answer is option (a)
Maximum covalencyless melting point.
Q.10. The above phenomenon can be explained as the basis of
(a) ionic potential of metal
(b) electron affinity
(c) ionization potential
(d) none of these
Correct Answer is option (a)
Ionic potential = charge/radius .
Q.11. The smallest ion among the following is
(a) Na+
(b) Al3+
(c) Mg2+
(d) Si4+
Correct Answer is option (d)
The isoelectronic ions Si4+, has the smallest size due to maximum nuclear charge.
Q.12. Highest covalent character is found in which of the following
(a) CaF2
(b) CaCl2
(c) Cal2
(d) CaBr2
Correct Answer is option (c)
I– is highly polarisable.
Q.13. In the transition of Cu to Cu2+, there is a decrease in
(a) Atomic number
(b) Atomic mass
(c) Equivalent weight
(d) Number of valency electrons
Correct Answer is option (d)
No. of valency electrons will reduced.
Q.14. Which of the order for ionisation energy is correct
(a) Be > B > C > N > O
(b) B < Be < C < O < N
(c) B < Be < C < N < O
(d) B < Be < N < C < O
Correct Answer is option (b)
Ionisation energy increases across the period but due to stable half filled configuration of VA group, its I.E. is more than VI-A group.
Q.15. The correct order of second ionization potential of carbon, nitrogen, oxygen and fluorine is
(a) C > N > O > F
(b) O > N > F > C
(c) O > F > N > C
(d) F > O > N > C
Correct Answer is option (d)
The ionization potential increases across the period but the second ionization potential of oxygen is highest among them because after the removal of 1 e- the 2 e- is to be removed from half-filled orbital which is difficult.
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