Chemistry Test 185

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Chemistry Test 185

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Weekly Quiz Competition

Question 1
4 / -1

The statement that is not correct for the periodic classification of elements is :

A
the properties of elements are the periodic functions of their atomic numbers.

B
non–metallic elements are lesser in number than metallic elements.

C
the first ionisation energies of elements along a period do not vary in a regular manner with increase in atomic number.

D
for transition elements the d–subshells are filled with electrons monotonically with increase in atomic number.

Solution

The d-sub shells are not filled with electrons monotonically with increase in atomic number. There are some exceptions like Cr, Cu etc.
Question 2
4 / -1

The increasing order of acidic nature of Li₂O, BeO, B₂O₃, CuO is

A
Li₂O < BeO < CuO < B₂O₃

B
BeO < CuO < B₂O₃ < Li_2O

C
Li₂O < CuO < BeO < B₂O₃

D
B₂O₃ < CuO < BeO < Li_2O

Solution

On moving left to right in periodic table, acidic nature of oxide increases. So, metallic oxides are basic, non metallic oxides are acidic and BeO is amphoteric.
Question 3
4 / -1

Which of the following statements is wrong for the transition elements ?

A
Transition elements are placed from 3^rd to 6^th period.

B
Last electron enters in (n – 1) d orbital.

C
Exhibits variable valency.

D
General electronic configuration is (n – 1) d^1–10 ns^0–2

Solution

Transition elements starts from 4^th period as they have valence shell configuration (n – 1)d^{1 - 10} ns^0–2. Thus for n = 3, there is no d-orbital available.
Question 4
4 / -1

Which of the following is not isoelectronic series ?

A
Cl^-,P^3-,Ar

B
N^3- ,Ne ,Mg²⁺

C
B³⁺,He ,Li⁺

D
N^3-,S^2- ,Cl^-

Solution

The number of electrons present are not same in (D) options N^3– = 10 electrons, S^2– = 18 electrons, Cl^– = 18 electrons. So this group does not represent the isoelectronic species.
Question 5
4 / -1

The first ionisation energy of Al is smaller than that of Mg because :

A
the atomic number of Al is greater than that of Mg.

B
the atomic size of Al is less than that of Mg.

C
penetration of s-subshell electrons in case of Mg is greater than that of p-subshell in Al

D
Mg has incompletely filled s-orbital.

Solution

Penetration of p-subshell electron is less than s-subshell electrons. In case of Mg, the first electron is to be removed from completely filled 3s² valence shell configuration as compared to partially filled 3p¹ of Al. These two factors collectively accounts for the higher ionisation energy of Mg than that of Al. Therefor, (C) option is correct.
Question 6
4 / -1

Which of the following properties is affected by stable configuration of an atom ?

(a) Electronegativity

(b) Ionisation potential

(c) Electron affinity

Correct answer is :

A
only electronegativity

B
only ionisation energy

C
electron affinity and ionsation energy both

D
all of the above

Solution

The addition of extra electron is difficult to the atom having stable configuration and so electron affinity will be less or zero. Similarly the removal of electron is quite difficult for stable configuration and so ionisation energy is higher. Hence electron affinity and ionisation energy both are affected by stable configuration.
Question 7
4 / -1

Fluorine has the highest electronegativity among the ns² np⁵ group on the Pauling scale, but the electron affinity of fluorine is less than that of chlorine because :

A
the atomic number of fluorine is less than that of chlorine.

B
fluorine being the first member of the family behaves in an unusual manner.

C
chlorine can accommodate an electron better than fluorine by utilising its vacant 3d–orbital.

D
small size, high electron density and an increased electron repulsion makes addition of an electron to fluorine less favourable than that in the case of chlorine.

Solution

There is more interelectronic repulsion in 2p-subshell of fluorine than 3p-subshell of chlorine due to its small atomic size. So extra electron will be added easily in 3p-subshell of chlorine as compared to 2p-subshell of fluorine.
Question 8
4 / -1

Which of the following orders is incorrect ?

A
O^2- < F^- < Na⁺ < Mg²⁺ ; Increasing Z_effective

B
Mg²⁺ < Na⁺ < F^- < O^2-Increasing size

C
O^2- < F^- < Na⁺ < Mg²⁺ ;Increasing ionisation energy

D
O^2- < F^- < Na⁺ < Mg²⁺ ; Increasing electron affinity

Solution

Option A is incorrect.

All are isoelectric species so with increasing effective nuclear charge :

(i) size decreases (ionic size ∝ 1/nuclear charge).

(ii) ionisation energy increases as size decreases and nuclear charge increases.

(iii) electron affinity increases as size decreases and nuclear charge increases.
Question 9
4 / -1

The correct order of the increasing ionic character is :

A

B

C

D

Solution

According to Fajan's rule as size of cations increase, their polarising power decrease and thus ionic character increase. So option (A) is correct.
Question 10
4 / -1

In which of the following sets the central atom of each member involves sp³ hybridisation ?

A

B

C

D

Solution

Hybridisation is calculated by steric number rule.

Steric no. = no. of atoms attached to central atom + no. of lone pairs present on central atom.

Steric no. for IO₄^- = 4 + 0 = 4 sp³ ; Steric no. for ICI₄^- = 4 + 2 = 6 sp³d²

Steric no. for IF₄⁺ = 4 + 1 = 5 sp³d ; Steric no. for SO₃ = 3 + 0 = 3 sp²

Steric no. for SO₄^2- = 4 + 0 = 4 sp3 ; Steric no. for SO₃^2- = 3 + 1 = 4 sp³

Steric no. for PCI₄⁺ = 4 + 0 = 4 sp³ ; Steric no. for BF₄^- = 4 + 0 = 4 sp³

Steric no. for XeO₃ = 3 + 1 = 4 sp³ ; Steric no. for XeO₄ = 4 + 0 = 4 sp³

Steric no. for XeF₄ = 4 + 2 = 6 sp³d² ; Steric no. for CIO₄^- = 4 + 0 = 4 sp³