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Explore popular questions from Classification of Elements and Periodicity in Properties for JEE Advanced. This collection covers Classification of Elements and Periodicity in Properties previous year JEE Advanced questions hand picked by experienced teachers.

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Classification of Elements and Periodicity in Properties

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Q 1. The correct order of second ionisation potential of carbon,nitrogen, oxygen and fluorine is

A

{tex} \mathrm { C } > \mathrm { N } > \mathrm { O } > \mathrm { F } {/tex}

B

{tex} \mathrm { O } > \mathrm { N } > \mathrm { F } > \mathrm { C } {/tex}

{tex} \mathrm { O } > \mathrm { F } > \mathrm { N } > \mathrm { C } {/tex}

D

{tex} \mathrm { F } > \mathrm { O } > \mathrm { N } > \mathrm { C } {/tex}

Explanation



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Q 2. The element with the highest first ionization potential is

A

boron

B

carbon

nitrogen

D

oxygen

Explanation

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Q 3. Which one of the following is the smallest in size?

A

{tex} \mathrm { N } ^ { 3 -} {/tex}

B

{tex} \mathrm{O ^ { 2- } }{/tex}

C

{tex} \mathrm { F } ^ { - } {/tex}

{tex} \mathrm { Na } ^ { + } {/tex}

Explanation

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Q 4. Amongst the following elements (whose electronic configurations are given below), the one having the highest ionization energy is:

A

{tex} [ \mathrm {{ Ne } ] 3 s ^ { 2 } 3 p ^ { 1 }} {/tex}

{tex} [ \mathrm {{ Ne } ] 3 s ^ { 2 } 3 p ^ { 3 }} {/tex}

C

{tex} [ \mathrm {{ Ne } ] 3 s ^ { 2 } 3 p ^ { 2 }} {/tex}

D

{tex} [ \mathrm {{ Ne } ] 3 d ^ { 10 } 4 s ^ { 2 } 4 p ^ { 3 }} {/tex}

Explanation

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Q 5. The correct order of acidic strength is

{tex} \mathrm { Cl } _ { 2 } \mathrm { O } _ { 7 } > \mathrm { SO } _ { 2 } > \mathrm { P } _ { 4 } \mathrm { O } _ { 10 } {/tex}

B

{tex} \mathrm { CO } _ { 2 } > \mathrm { N } _ { 2 } \mathrm { O } _ { 5 } > \mathrm { SO } _ { 3 } {/tex}

C

{tex} \mathrm { Na } _ { 2 } \mathrm { O } > \mathrm { MgO } > \mathrm { Al } _ { 2 } \mathrm { O } _ { 3 } {/tex}

D

{tex} \mathrm { K } _ { 2 } \mathrm { O } > \mathrm { CaO } > \mathrm { MgO } {/tex}

Explanation

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Q 6. Identify the correct order of acidic strengths of {tex} \mathrm { CO } _ { 2 } , \mathrm { CuO } {/tex},
{tex} \mathrm { CaO } , \mathrm { H } _ { 2 } \mathrm { O } {/tex}

{tex} \mathrm { CaO } < \mathrm { CuO } <\mathrm { H } _ { 2 } \mathrm { O } < \mathrm { CO } _ { 2 } {/tex}

B

{tex} \mathrm { H } _ { 2 } \mathrm { O } < \mathrm { CuO } < \mathrm { CaO } < \mathrm { CO } _ { 2 } {/tex}

C

{tex} \mathrm { CaO } < \mathrm { H } _ { 2 } \mathrm { O } < \mathrm { CuO } < \mathrm { CO } _ { 2 } {/tex}

D

{tex} \mathrm { H } _ { 2 } \mathrm { O } < \mathrm { CO } _ { 2 } < \mathrm { CaO } < \mathrm { CuO } {/tex}

Explanation

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Q 7.

Which of the following has the electronic configuration ?

A

Cr

C

D

V

Explanation

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Q 8.

Electron affinity is positive when

A

is formed from

is formed from

C

is formed from

D

Electron affinity is always a negative value

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Q 9.

are elements in the third short period. Oxide of is ionic, that of is amphoteric and of a giant molecule. and will have atomic number in the order

B

C

D

Explanation

Thus, atomic number increases in the order

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Q 10.

First long period contains …….. elements

A

8

18

C

32

D

2

Explanation

First long period starts with 3rd period [K(19) Kr(36)]

Thus, total = 18 elements

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Q 11.

Li resemble Mg. This is called diagonal relationship which is attributed to

Same value of electronegativity

B

Same value of electron affinity

C

Penetration of sub-shells

D

Identical effective nuclear charge

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Q 12.

Which of the following does not represent the correct order of the properties indicated?

(size)

B

(size)

C

(unpaired electron)

D

(unpaired electron)

Explanation

is smallest in size

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Q 13.

If Aufbau and Hund’s rule are not used, then incorrect statement is

would be coloured ion

B

Na will be in same -block (if these rules are true)

C

Cu would be -block element

D

Magnetic moment of Cr(24) would be zero

Explanation

Colourless ion due to lack of electron in -orbitals

Thus, (a) is incorrect

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Q 14.

Representative elements belong to

- and -block

B

-block

C

- and -block

D

-block

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Q 15.

Which of the following oxides is highly basic?

A

B

C

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Q 16.

If Aufbau rule is not followed, K-19 will be placed in ……. block

A

B

D

Explanation

E.C. of K = 19 in the absence of Aufbau rule is

Last-filling electron goes into -orbital

Thus, -block element

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Q 17.

Which of the following metals forms a amphoteric oxide?

A

Ca

B

Ni

Zn

D

Fe

Explanation

Basic oxide

Acidic oxide

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Q 18.

Catenation properties of C, Si, are in order

B

C

D

None of the above is correct

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Q 19.

Select the correct statement

More active metals are on the left side of the Periodic Table

B

Less active metals are on the left side of the Periodic Table

C

Reducing power decreases moving down the group

D

All the above are correct statements

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Q 20.

The electronegativities of elements and are 1.2 and 3.4 units respectively. The type of bond connecting and in compound is

A

Covalent

Ionic

C

Coordinate covalent

D

Polar covalent

Explanation

is electropositive element

is electronegative element

Thus, ionic bond is formed

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Q 21.

Which of the following will have maximum electron affinity?

A

B

D

Explanation

(b) and (d) with configuration has zero values of EA

(c) with empty -orbital has greater EA than F(a)

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Q 22.

Select the correct alternate

A

Due to lanthanide contraction and have almost equal size

B

Due to completion of -subshell, the electronic charge density in this subshell becomes very high which increases the inter-electronic repulsion hence, size increases

Both (a) & (b) are correct statements

D

Both of the above are incorrect statements

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Q 23.

Point out the oxide which is amphoteric in nature

A

CO

B

D

Explanation

is soluble in an acidic oxide

is soluble in a basic acid

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Q 24.

Covalent radius of nitrogen is 70 pm. Hence, covalent radius of boron is about

A

60 pm

110 pm

C

50 pm

D

40 pm

Explanation

Radius decreases along a period left to right

Thus, covalent radius of B would be higher than that of N

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Q 25.

Melting points of and will be in order

B

C

D

Explanation

Smaller the size of anion, smaller the polarizing power and thus larger the ionic character of . Thus, ionic character

and mp :