Periodic Table & Periodicity Questions — NEET Chemistry

Q1

The correct order of the atomic radii of C, Cs, Al, and S is :

A S < C < Al < Cs

B S < C < Cs < Al

C C < S < Al < Cs

D C < S < Cs < Al

Correct Answer

Option C

Solution

Atomic radii increase by moving down the group and decrease across a period.

Hence, the correct order of atomic radii is : C < S < Al < Cs.

Q2

The correct order of increasing metallic character of

\mathrm{Na}, \mathrm{Be}, \mathrm{P}, \mathrm{Mg}

and Si is

A $\mathrm{P}

B $\mathrm{P}

C $\mathrm{P}

D $\mathrm{Be}

Correct Answer

Option A

Solution

On moving left to right in a period, metallic character decreases and on moving top to bottom in a group, metallic character increases.

So, the correct order of increasing metallic character is

\mathrm{P}

\text { Electronegativity } \propto \frac{1}{\text { Metallic character }}

\begin{array}{llllll} & \mathrm{Na} & \mathrm{Mg} & \mathrm{Be} & \mathrm{Si} & \mathrm{P} \\ \text { Electronegativity (On Pauling scale) } & 0.9 & 1.2 & 1.5 & 1.8 & 2.1 \end{array} $$

Q3

Identify the incorrect statement from the following :

A The largest and the smallest species among

\mathrm{Mg}, \mathrm{Mg}^{2+}, \mathrm{Al}

and

\mathrm{Al}^{3+}

are Al and

\mathrm{Mg}^{2+}

respectively.

B The IUPAC name of the element with atomic number 107 is Unnilseptium.

C The similarity in behaviour of Li with Mg is referred to as 'diagonal relationship'

D The oxidation state and covalency of Al in

\left[\mathrm{AlCl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}

are 3 and 6, respectively.

Correct Answer

Option A

Solution

The largest species is Mg and the smallest one is $\mathrm{Al}^{3+}$ among $\mathrm{Mg}, \mathrm{Mg}^{2+}, \mathrm{Al}$ and $\mathrm{Al}^{3+}$ .

Unnilseptium element has atomic number 107 $\mathrm{Li}$ and Mg are diagonally related and have similar properties.

The oxidation state and covalency of Al in $\left[\mathrm{AlCl}\left(\mathrm{H}_2 \mathrm{O}\right)_5\right]^{2+}$ are +3 and 6 respectively ∴ Statement A is incorrect

Q4

The correct decreasing order of atomic radii (pm) of

\mathrm{Li}, \mathrm{Be}, \mathrm{B}

and

\mathrm{C}

is

A

\mathrm{Be}>\mathrm{Li}>\mathrm{B}>\mathrm{C}

B

\mathrm{Li}>\mathrm{Be}>\mathrm{B}>\mathrm{C}

C

\mathrm{C}>\mathrm{B}>\mathrm{Be}>\mathrm{Li}

D

\mathrm{Li}>\mathrm{C}>\mathrm{Be}>\mathrm{B}

Correct Answer

Option B

Solution

As the atomic number in a period increases, the effective nuclear charge also increases hence, atomic radii along the period decreases.

Correct order of atomic radii

\mathrm{Li}>\mathrm{Be}>\mathrm{B}>\mathrm{C}

Q5

Which of the following is correctly matched?

A Basic oxides

\Rightarrow \mathrm{In}_2 \mathrm{O}_3, \mathrm{~K}_2 \mathrm{O}, \mathrm{SnO}_2

B Neutral oxides

\Rightarrow \mathrm{CO}, \mathrm{NO}_2, \mathrm{~N}_2 \mathrm{O}

C Acidic oxides

\Rightarrow \mathrm{Mn}_2 \mathrm{O}_7, \mathrm{SO}_2, \mathrm{TeO}_3

D Amphoteric oxides

\Rightarrow \mathrm{BeO}, \mathrm{Ga}_2 \mathrm{O}_3, \mathrm{GeO}

Correct Answer

Option C

Solution

\mathrm{Mn}_2 \mathrm{O}_7, \mathrm{SO}_2, \mathrm{TeO}_3

are acidic oxides.

Q6

The correct order of first ionization enthalpy for the given four elements is :

A C < F < N < O

B C < N < F < O

C C < N < O < F

D C < O < N < F

Correct Answer

Option D

Solution

$\bullet$ Generally, on moving left to right in a period.

First ionization enthalpy of elements increases due to increase in effective nuclear charge.

$\bullet$ Due to more stable half-filled outer electronic configuration (2s 2 2p 3 ) of N, its first ionization enthalpy is more than O.

So, correct order of IP is : C

Q7

Decrease in size from left to right in actinoid series is greater and gradual than that in lanthanoid series due to

A 5f orbitals have greater shielding effect

B 4f orbitals are penultimate

C 4f orbitals have greater shielding effect

D 5f orbitals have poor shielding effect

Correct Answer

Option D

Solution

Due to more diffused nature of 5f orbitals as compared to 4f orbitals the shielding effect of 5f is poor, resulting in the decrease in size from left to right in actinoid series which is greater and gradual than that in lanthanoid series.

Q8

Fluorine is a stronger oxidising agent than chlorine because : (a) F-F bond has a low enthalpy of dissociation. (b) Fluoride ion (F

-

) has high hydration enthalpy. (c) Electron gain enthalpy of fluorine is less negative than chlorine. (d) Fluorine has a very small size. Choose the most appropriate answer from the options given :

A (b) and (c) only

B (a) and (b) only

C (a) and (c) only

D (a) and (d) only

Correct Answer

Option B

Solution

Fluorine is a stronger oxidising agent than chlorine due to (i) Low dissociation enthalpy of F-F bond (ii) High hydration enthalpy of F $-$ ion

Q9

If first ionization enthalpies of elements X and Y are 419 kJ mol

-

1 and 590 kJ mol

-

1 , respectively and second ionization enthalpies of X and Y are 3069 kJ mol

-

1 and 1145 kJ mol

-

1 , respectively. Then correct statement is :

A Both X and Y are alkaline earth metals

B X is an alkali metal and Y is an alkaline earth metal

C X is an alkaline earth metal and Y is an alkali metal

D Both X and Y are alkali metals

Correct Answer

Option B

Solution

As it can be observed from given data of question, in case of element 'X' there is huge difference between IP 1 and IP 2 hence it will have one electron in outermost shell and will be alkali metal.

While for 'Y' difference is not that high hence it will be alkaline earth metal.

Q10

The IUPAC name of an element with atomic number 119 is

A ununennium

B unnilennium

C unununnium

D ununoctium

Correct Answer

Option A

Solution

IUPAC name of element : 119 : ununennium