The s-Block Elements Test

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The s-Block Elements Test - 4

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Weekly Quiz Competition

Question 1
1 / -0

What is the oxidation state of K in KO₂?

A
+1

B
+3

C
+2

D
+4

Solution

K has +1 oxidation state. "The potassium atom losses one electron when forming compound as it has one electron on it's valence shell".
Question 2
1 / -0

2M + 2H₂O →2M⁺+ 2OH⁻+ H₂ What is M?

A
transition elements

B
alkaline earth metals.

C
noble gases

D
alkali metals.

Solution

M is alkali metal. Alkali metals reacts with water vigorously to form hydroxide (OH^-) and dihydrogen (H₂)
Question 3
1 / -0

Lithium has most negative E⁰ value, its reaction with water is less vigorous than that of sodium which has the least negative E⁰ value among the alkali metals. The reason is

A
it is a proton donor

B
it is a strong reducing agent

C
small size of Li

D
low hydration energy of Li

Solution

This behaviour of lithium is attributed to its small size and very high hydration energy. Li⁺ has maximum degree of hydration.
Question 4
1 / -0

The alkali metals dissolve in liquid ammonia giving deep blue solutions. When they become diamagnetic, the solution turns into

A
white colour

B
bronze colour

C
purple colour

D
green colour

Solution

When the solution become diamagnetic due to formation of electron cluster in which ammoniated electron with opposite spin group- it changes its color to bronze.
Question 5
1 / -0

Lithium halides are somewhat covalent because

A
high polarisation capability of lithium ion

B
Li+ ion is very large in size and has high tendency to distort electron

C
low polarisation capability of lithium ion

D
Li+ ion is very small in size and has low tendency to distort electron

Solution

Li⁺ ion has small size and maximum tendency to attract the electrons towards itself from the negative ion. This distortion of the electron cloud of the negative ion by the positive ion is known as polarization. Thus Li⁺has high polarisation capability. Lithium iodide is the most covalent in nature.
Question 6
1 / -0

Choose the correct trend of the melting and boiling points in the alkali metal halides, MF, MCl, MI, MBr.

A
MCl > MBr > MI

B
MF > MCl > MBr > MI

C
MI

D
MBr > MI

Solution

Smaller is the size of anion less is the polarizing power of anion so greater is the ionic character hence greater will be melting point.
Question 7
1 / -0

All alkali metal halides are soluble in water except

A
lithium

B
Francium

C
rubidium

D
caesium

Solution

Greater the lattice energy, higher is the melting point of the MX (metal halide) and lower is its solubility
Question 8
1 / -0

Why KO₂ is paramagnetic ?

A
because of lattice energy effects

B
because of large size of the metal ion

C
because of oxygen atom in -1/2 oxidation state

D
because of one unpaired electron in π^∗ 1p

Solution

KO₂ is a superoxide, K is in +1 oxidation state and O₂ has -1 oxidation state, thus each O has -1/2 oxidation state. So due to the presence of unpaired electron it behaves as free radical. As we know that a molecule containing unpaired electron is paramagnetic in nature, so KO₂ is paramagnetic in nature because of one unpaired electron in π*2p molecular orbital.
Question 9
1 / -0

Alkali metal chlorides donot form hydrates except for one metal. Choose from among the options given below.

A
Francium

B
caesium

C
lithium

D
rubidium

Solution

Lithium due to its high hydration energy(deliquescent) forms hydrate.

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Re-Attempt Weekly Quiz Competition

The s-Block Elements Test - 4

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The s-Block Elements Test - 4

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Question 1

+1

+3

+2

+4

Solution

Question 2

transition elements

alkaline earth metals.

noble gases

alkali metals.

Solution

Question 3

it is a proton donor

it is a strong reducing agent

small size of Li

low hydration energy of Li

Solution

Question 4

white colour

bronze colour

purple colour

green colour

Solution

Question 5

high polarisation capability of lithium ion

Li+ ion is very large in size and has high tendency to distort electron

low polarisation capability of lithium ion

Li+ ion is very small in size and has low tendency to distort electron

Solution

Question 6

MCl > MBr > MI

MF > MCl > MBr > MI

MI

MBr > MI

Solution

Question 7

lithium

Francium

rubidium

caesium

Solution

Question 8

because of lattice energy effects

because of large size of the metal ion

because of oxygen atom in -1/2 oxidation state

because of one unpaired electron in π∗ 1p

Solution

Question 9

Francium

caesium

lithium

rubidium

Solution

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because of one unpaired electron in π^∗ 1p