D- and F-Block Elements Questions for JEE Main 2027 with Solutions

Weightage of D- and F- Block Elements in JEE Main 2027

Understanding the D- and F-Block Elements weightage in JEE Main 2026 can help students prioritise this chapter during Chemistry preparation. The table below highlights the weightage to help you plan revision and focus on important topics.

Also Read: JEE Main Chemistry: 20 Most Repeated Topic-Based Questions 

D- and F- Block Elements JEE Main Questions

Question 1: The first transition series metal ' M ' has the highest enthalpy of atomisation in its series. One of its aquated ions $\left(\mathrm{M}^{\mathrm{n}}\right)$ exists in green colour. The nature of the oxide formed by the above $\mathrm{M}^{\mathrm{n}-}$ ion is :

1) neutral
2) acidic
3) basic
4) amphoteric

Correct answer: (3)

Solution:

* In 3d series Vanadium has highest enthalpy of atomization and colour of $\mathrm{V}^{+3}$ is green.
* Oxide form by $\mathrm{V}^{+3}$ is $\mathrm{V}_2 \mathrm{O}_3$ (Basic oxide)

Hence, the correct answer is option (3).

Question 2: The number of unpaired electrons responsible for the paramagnetic nature of the following complex species are respectively: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-},\left[\mathrm{FeF}_6\right]^{3-},\left[\mathrm{CoF}_6\right]^{3-},\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$

1) 1,5,4,2
2) 1,5,5,2
3) 1,1,4,2
4) 1,4,4,2

Correct answer: (1)

Solution:

$\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{3-} \Rightarrow \mathrm{Fe}^{3+}, \mathrm{d}^5, \mathrm{t}_{2 \mathrm{~g}}^5 \mathrm{e}_{\mathrm{g}}^0$

                       1 unpaired electron

$\left[\mathrm{FeF}_6\right]^{3-} \Rightarrow \mathrm{Fe}^{3+}, \mathrm{d}^5, \mathrm{t}_{2 \mathrm{~g}}^3 \mathrm{e}_{\mathrm{g}}^2 \Rightarrow 5$ unpaired electrons $\left[\mathrm{CoF}_6\right]^{3-} \Rightarrow \mathrm{Co}^{3+}, \mathrm{d}^6, \mathrm{t}_{2 \mathrm{~g}}^4 \mathrm{e}_{\mathrm{g}}^2 \Rightarrow 4$ unpaired electrons

$\begin{aligned} {\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-} \Rightarrow } & \mathrm{Mn}^{3+}, \mathrm{d}^4,\end{aligned} \mathrm{t}_{2 \mathrm{~g}}^4 \mathrm{e}_{\mathrm{g}}^0 \mathrm{C}$.

                               2 unpaired electron

Hence, the correct answer is option (1).

Question 3: The correct set of ions (aqueous solution) with the same colour from the following is :

1) $\mathrm{V}^{2+}, \mathrm{Cr}^{3+}, \mathrm{Mn}^{3+}$
2) $\mathrm{Zn}^{2+}, \mathrm{V}^{3+}, \mathrm{Fe}^{3+}$
 
3) $\mathrm{Ti}^{4+}, \mathrm{V}^{4+}, \mathrm{Mn}^{2+}$
 
4) $\mathrm{Sc}^{3+}, \mathrm{Ti}^{3+}, \mathrm{Cr}^{2+}$

Correct answer: (1)

Solution:

(1) $\mathrm{V}^{2+}$ (Violet), Cr3+ (Violet), $\mathrm{Mn}^{3+}$ (Violet)
(2) Zn2+ (Colourless), V3+ (Green), Fe3+ (Yellow)
(3) Ti4+ (Colourless), V4+ (Blue), $\mathrm{Mn}^{2+}$ (Pink)
(4) $\mathrm{Sc}^{3+}$ (Colourless), $\mathrm{Ti}^{3+}$ (Purple), $\mathrm{Cr}^{2+}$ (Blue)

Hence, the correct answer is option (1).

Question 4: Lanthanoid ions with 4f7 configuration are :
(A) $\mathrm{Eu}^{2+}$
(B) $\mathrm{Gd}^{3+}$
(C) $\mathrm{Eu}^{3+}$
(D) $\mathrm{Tb}^{3+}$
(E) $\mathrm{Sm}^{2+}$ Choose the correct answer from the options given below :

1) (A) and (B) only
 
2) (A) and (D) only
 
3) (B) and (E) only
 
4) (B) and (C) only

Correct answer: (1)

Solution:

$\begin{aligned} & { }_{63} \mathrm{Eu}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^7 6 \mathrm{~s}^0 \\ & { }_{64} \mathrm{Gd}^{3+}-[\mathrm{Xe}] 4 \mathrm{f}^7 5 \mathrm{~d}^0 6 \mathrm{~s}^0 \\ & { }_{63} \mathrm{Eu}^{3+}-[\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^0 \\ & { }_{65} \mathrm{~Tb}^{3+}-[\mathrm{Xe}] 4 \mathrm{f}^8 6 \mathrm{~s}^0 \\ & { }_{62} \mathrm{Sm}^{2+}-[\mathrm{Xe}] 4 \mathrm{f}^6 6 \mathrm{~s}^0 \\ & \mathrm{Eu}^{2+} \& \mathrm{Gd}^{3+}\end{aligned}$

Hence, the correct answer is option (1).

Question 5: Given below are two statements :
Statement I: In the oxalic acid vs KMnO4 (in the presence of dil H2SO4 ) titration, the solution needs to be heated initially to 60C, but no heating is required in Ferrous ammonium sulphate (FAS) vs KMnO4 titration (in the presence of dil H2SO4 ). Statement I: In oxalic acid vs KMnO4 titration, the initial formation of MnSO4 takes place at high temperature, which then acts as a catalyst for further reaction. In the case of FAS vs KMnO4, heating oxidises Fe2+ into Fe3- by oxygen of air, and error may be introduced in the experiment. In light of the above statements, choose the correct answer from the options given below :

1) Statement I is false, but Statement II is true 
2) Both Statement I and Statement II are true
3) Statement I is true,e but Statement II is false
 
4) Both Statement I and Statement II are false

Correct answer: (2)

Solution:

$\begin{aligned} & 2 \mathrm{MnO}_4^{-}+5(\mathrm{COO})_2^{2-}+16 \mathrm{H}^{+} \rightarrow &  10 \mathrm{CO}_2+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O}\end{aligned}$

This reaction is slow at room temperature but becomes fast at 60C. Manganese(II) ions catalyse the reaction; thus, the reaction is autocatalytic; once manganese(II) ions are formed, it becomes faster and faster.

The titration of FAS v/sKMnO4 does not require heating because at higher temperatures the oxidation of Fe+2 to Fe+3 by atmospheric O2 will be prominent.

Hence, the correct answer is option (2).

Question 6: The incorrect relationship in the following pairs in relation to ionisation enthalpies is :

1) Mn+<Cr+
2) Mn+<Mn2+
3) Fe2+<Fe3+
4) Mn2+<Fe2+

Correct answer: (4)

Solution:

Mn2+:[Ar]3~d5

Half-filled stability
Therefore, more I.E. than Fe2+

Fe2+:[Ar]3~d6

Hence, the correct answer is option (4). 

Question 7: The metal ions that have the calculated spin-only magnetic moment value of 4.9 B.M. are
A. Cr2+
B. Fe2+
C. Fe3+
D. Co2+
E. Mn3+ Choose the correct answer from the options given below

1) A, C and E only
2) A, D and E only
 
3) B and E only
4) A, B and E only

Correct answer: (4)

Solution:

Given magnetic moment =4.9~B.M.
We know M.M =n(n+2) B.M.
Where, n No. of unpaired e-

4.9=n(n+2)

Solving this equation, we find that n=4.

(A) 24Cr2+[Ar]3~d4 (4 unpaired e-)
(B) 26Fe2+[Ar]3~d6 (4 unpaired e-)
(C) 26Fe3+[Ar]3~d5 (5 unpaired e-)
(D) ${ }_{27} \mathrm{Co}^{2+} \Rightarrow[\mathrm{Ar}] 3 \mathrm{~d}^7 \quad\left(3\right.$ unpaired $\left.\mathrm{e}^{-}\right)$
(E) 25Mn3+[Ar]3~d4(4 unpaired e )

The metal ions with a spin-only magnetic moment of approximately 4.9 B.M. are Cr2+,Fe2+and Mn3+. Because the value of n=4 for these ions.

Hence, the correct answer is option (4).

Question 8: Which of the following ions is the strongest oxidising agent?
(Atomic Number of Ce=58,Eu=63,~Tb=65,Lu=71 ]

1) Lu3+
2) Eu2+
3) Tb4+
4) Ce3+

Correct answer: (3)

Solution:

Tb4+ is the strongest oxidising agent

Hence, the correct answer is option (3).

Question 9: Preparation of potassium permanganate from MnO2 involves a two-step process in which the $1^{\text {st }}$ step is a reaction with KOH and KNO3 to produce

1) K4Mn(OH)6
2) K3MnO4
3) KMnO4
4) K2MnO4

Correct answer: (4)

Solution:

$
\mathrm{MnO}_2 \xrightarrow[\mathrm{KNO}_3, \mathrm{~A}]{\mathrm{KOH}} \mathrm{~K}_2 \mathrm{MnO}_4
$

Hence, the correct answer is option (4).

Question 10: Given below are two statements :
Statement I: CrO3 is a stronger oxidising agent than MoO3
Statement II: Cr(VI) is more stable than Mo(VI)
In light of the above statements, choose the correct answer from the options given below

1) Statement I is false, but Statement II is true
2) Statement I is true,e but Statement II is false
3) Both Statement I and Statement II are true
4) Both Statement I and Statement II are false

Correct answer: (2)

Solution: Statement I is true, but Statement II is false.
Cr(VI) is less stable than Mo(VI)

Stability : Cr6+<Mo6+
Hence, CrO3 is easily reduced to Cr-3 as compared to MoO3 and shows a stronger oxidising nature.

Hence, the correct answer is option (2).

Question 11: Match List - I with List - II.

Choose the correct answer from the options given below :

1) (A)-(IV), (B)-(II), (C)-(III), (D)-(I)
 
2) (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
 
3) (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
 
4) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct answer: (2)

Solution:

Bronze Cu,Sn
Brass Cu,Zn
UK silver coin Cu,Ni
Stainless steel Fe,Cr,Ni,C

Hence, the correct answer is option (2).

Question 12: The correct decreasing order of spin-only magnetic moment values (BM) of Cu+,Cu2+,Cr2+ and Cr3+ ions is:

1) Cu+>Cu2+>Cr3+>Cr2+
2) Cu2+>Cu+>Cr2+>Cr3+
3) Cr2+>Cr3+>Cu2+>Cu+
4) Cr3+>Cr2+>Cu+>Cu2+

Correct answer: (3)

Solution:

Cu+3~d10  0BM

Cu2+3~d9   1.73BM

Cr2+3~d4 4.89BM  

Cr3+3~d3  3.87BM 

So order :

Cr2+>Cr3+>Cu2+>Cu+

Hence, the correct answer is option (3).

"Question 13: Among $\mathrm{Sc}, \mathrm{Mn}, \mathrm{Co}$ and Cu, identify the element with the highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ___________ BM (in nearest integer).

1) 3.87
2) -
3) -
4) -

Correct answer: (1)

Solution:

Enthalpy of atomisation of : Sc = $326 \mathrm{~kJ} / \mathrm{mol}, \mathrm{Mn}=$ $281 \mathrm{~kJ} / \mathrm{mol}, \mathrm{Co}=425 \mathrm{~kJ} / \mathrm{mol}, \mathrm{Cu}=338 \mathrm{~kJ} / \mathrm{mol}$

$ \mathrm{Co}^{2+}:[\mathrm{Ar}] 4 \mathrm{~s}_0 3 \mathrm{~d}^7: $ 

since the number of unpaired electrons is 3. Therefore, the magnetic moment will be,

 =3(3+2)=15=3.87

Hence, the answer is 3.87.

Question 14: KMnO4 acts as an oxidising agent in acidic medium. ‘X’ is the difference between the oxidation states of Mn in reactant and product. ‘Y’ is the number of ‘d’ electrons present in the brown-red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y is _________.

1) 10
2) -
3) -
4) -

Correct answer: (1)

Solution:

In acidic medium, $\mathrm{KMnO}_4$ acts as an oxidizing agent

$
\mathrm{MnO}_4^{-} \rightarrow \mathrm{Mn}^{2+}
$

Oxidation state of Mn in $\mathrm{MnO}_4{ }^{-}=+7$
Oxidation state of Mn in $\mathrm{Mn}^{2+}=+2$

So,

$
X=7-2=5
$

$\begin{aligned} & 6 \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{Fe}^{3+}+\mathrm{H}_2 \mathrm{O}  \rightarrow\left[\mathrm{Fe}_3\left(\mathrm{OH}_2\right)\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+2 \mathrm{H}^{\oplus} \\ & {\left[\mathrm{Fe}_3(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right)_6\right]^{\oplus}+4 \mathrm{H}_2 \mathrm{O}}  \rightarrow \underset{\text { Brown red ppt }}{\left[\mathrm{Fe}(\mathrm{OH})_2\left(\mathrm{CH}_3 \mathrm{COO}\right]+\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}^{\oplus}\right.}\end{aligned}$

$\mathrm{Fe}^{3+} \Rightarrow 3 \mathrm{~d}^5 4 \mathrm{~s}^0$ contains 5 d electrons
So $\mathrm{Y}=5$

X+Y=5+5=10

Hence, the correct answer is 10.

Question 15: Choose the correct tests with respective observations.
(A) $\mathrm{CuSO}_4$ (acidified with acetic acid) + $\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow$ Chocolate brown precipitate.
(B) $\mathrm{FeCl}_3+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow$ Prussian blue precipitate.
(C) $\mathrm{ZnCl}_2+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$, neutralised with $\mathrm{NH}_4 \mathrm{OH}$ $\rightarrow$ White or bluish white precipitate.
(D) $\mathrm{MgCl}_2+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \rightarrow$ Blue precipitate.
(E) $\mathrm{BaCl}_2+\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]$, neutralised with NaOH $\rightarrow$ White precipitate.
Choose the correct answer from the options given below :

1) A, D and E only
2) B, D and E only
3) A, B and C only
4) C, D and E only

Correct answer: (3)

Solution:

$\begin{aligned}
&\begin{array}{r}
2 \mathrm{CuSO}_4+ \
\mathrm{K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \xrightarrow{\mathrm{CH}_3 \mathrm{COOH}} \\
\mathrm{Cu}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]+2 \mathrm{~K}_2 \mathrm{SO}_4
\end{array}\\
 &_{\text { (Chocolate brown ppt.) }}
\end{aligned}$

$\begin{aligned}
& 4 \mathrm{FeCl}_3+3 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \longrightarrow \\
& \mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3+12 \mathrm{KCl} \\
&_{ \text { (Prussian Blue ppt.) }} \\
& 3 \mathrm{ZnCl}_4+2 \mathrm{~K}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right] \xrightarrow{\mathrm{NH}_4 \mathrm{OH}} \\
& \mathrm{~K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2+6 \mathrm{KCl} \\
&_{ \text { (White or bluish white ppt.) }}
\end{aligned}$

Magnesium ion $\left(\mathrm{Mg}^{2+}\right)$ does not form a blue precipitate with potassium ferrocyanide.

$\mathrm{Mg}^{2+}$ does not form an insoluble salt with $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$

So, the solution remains clear.

$\mathrm{Ba}^{2+}$ does NOT form a definite or reliable precipitate with $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$
None or very faint turbidity occurs.

Hence, the correct answer is option (3).

Question 16: The spin only magnetic moment (  ) value (B.M.) of the compound with strongest oxidising power among $\mathrm{Mn}_2 \mathrm{O}_3, \mathrm{TiO}$ and VO is _______ B.M. (Nearest integer).

1) 5
2) -
3) -
4) -

Correct answer: (1)

Solution:

Strongest oxidising power among the option is $\mathrm{Mn}_2 \mathrm{O}_3$ because of $\mathrm{E}^{\circ}$ value

$\mathrm{E}_{\mathrm{Mn}^{+3} / \mathrm{Mn}^{+2}}^{\circ}=+1.57 \mathrm{~V}$

For $\mathrm{Mn}^{3+}$, the electronic configuration is d4. The spin-only magnetic moment () is calculated as follows:

$
\mu=\sqrt{n(n+2)} \text { вм }
$

Where n is the number of unpaired electrons. Here, n=4 :

$
\mu=\sqrt{4(4+2)}=\sqrt{24} \mathrm{BM}
$
This calculates to approximately 4.89 BM, which rounds to an integer value of 5 BM. 

Hence, the answer is 5. 

Question 17: The spin only magnetic moment (  ) value (B.M.) of the compound with strongest oxidising power among $\mathrm{Mn}_2 \mathrm{O}_3, \mathrm{TiO}$ and VO is _______ B.M. (Nearest integer).

1) 5
2) -
3) -
4) -

Correct answer: (1)

Solution:

Strongest oxidising power among the option is $\mathrm{Mn}_2 \mathrm{O}_3$ because of $\mathrm{E}^{\circ}$ value

$\mathrm{E}_{\mathrm{Mn}^{+3} / \mathrm{Mn}^{+2}}^{\circ}=+1.57 \mathrm{~V}$

For $\mathrm{Mn}^{3+}$, the electronic configuration is d4. The spin-only magnetic moment () is calculated as follows:

$
\mu=\sqrt{n(n+2)} \text { вм }
$

Where n is the number of unpaired electrons. Here, n=4 :

$
\mu=\sqrt{4(4+2)}=\sqrt{24} \mathrm{BM}
$
This calculates to approximately 4.89 BM, which rounds to an integer value of 5 BM. 

Hence, the answer is 5. 

Question 18: Pair of transition metal ions having the same number of unpaired electrons is :

1) $\mathrm{V}^{2+}, \mathrm{Co}^{2+}$
2) Ti2+,Co2+
3) Fe3+,Cr2+
4) $\mathrm{Ti}^{3+}, \mathrm{Mn}^{2+}$

Correct answer: (1)

Solution:

So $\mathrm{V}^{2+} \& \mathrm{Co}^{2+}$ same number of unpaired electrons.

Hence, the correct answer is option (1). 

Question 19: Consider ' n ' is the number of lone pairs of electrons present in the equatorial position of the most stable structure of $\mathrm{ClF}_3$. The ions from the following with ' n ' number of unpaired electrons are :
A. V3+
B. $\mathrm{Ti}^{3+}$
C. $\mathrm{Cu}^{2+}$
D. $\mathrm{Ni}^{2+}$
E. $\mathrm{Ti}^{2+}$ Choose the correct answer from the options given below :

1) A and C only
 
2) A, D and E only
3) B and C only
4) B and D only

Correct answer: (2)

Solution:

$\mathrm{ClF}_3$

$\mathrm{n}=2$ (No. of lone pairs present in the equatorial plane)
                                                         (Unpaired e )
(A) $\mathrm{V}^{+3}:[\mathrm{Ar}] 3 \mathrm{~d}^2$                          2

(B)$\mathrm{Ti}^{31}:[\mathrm{Ar}] 3 \mathrm{~d}^1$                          1

(C) $\mathrm{Cu}^{+2}:[\mathrm{Ar}] 3 \mathrm{~d}^9$                       1

(D) $\mathrm{Ni}^{12}:[\mathrm{Ar}] 3 \mathrm{~d}^8$                         2

(E)Ti+3:[Ar]3~d2                          2

Hence, the correct answer is option (2).

Question 20: The correct option with the order of melting points of the pairs $(\mathrm{Mn}, \mathrm{Fe}),(\mathrm{Tc}, \mathrm{Ru})$ and $(\mathrm{Re}, \mathrm{Os})$ is :

1) $\mathrm{Fe}<\mathrm{Mn}, \mathrm{Ru}<\mathrm{Tc}$ and $\mathrm{Re}<\mathrm{Os}$
 
2)
$\mathrm{Mn}<\mathrm{Fe}, \mathrm{Tc}<\mathrm{Ru}$ and $\mathrm{Re}<\mathrm{Os}$
 
3) $\mathrm{Mn}<\mathrm{Fe}, \mathrm{Tc}<\mathrm{Ru}$ and $\mathrm{Os}<\mathrm{Re}$
 
4)
$\mathrm{Fe}<\mathrm{Mn}, \mathrm{Ru}<\mathrm{Tc}$ and $\mathrm{Os}<\mathrm{Re}$

Correct answer: (3)

Solution:

Mn,TcRe are in the same group.
Also, $\mathrm{Fe}, \mathrm{Ru}, \mathrm{Os}$, are same group elements.

• Fe has a higher melting point than Mn,Mn<Fe. Mn has a weaker metallic bond due to its electronic configuration (half-filled configuration). The half-filled configuration (stable) results in less delocalisation of electrons and weaker attraction between atoms, requiring less energy to break apart and melt. But Fe has localised electrons and hence stronger metallic bonds, leading to a higher melting point.

• Tc has a lower melting point than $ \ mathrm {Ru} $, \mathrm{Tc}<\mathrm{Ru}$. This is because of the electronic configuration of its d5-orbitals.

• Re has a higher melting point than $\mathrm{Os}, \mathrm{Os}<\operatorname{Re}$. The higher melting point of Re is due to the strength of atomic bonding.

M.P. $\Rightarrow \mathrm{Mn}<\mathrm{Fe}, \mathrm{Tc}<\mathrm{Ru}, \mathrm{Os}<\mathrm{Re}$

Hence, the correct answer is option (3).

Question 21: The amphoteric oxide among $\mathrm{V}_2 \mathrm{O}_3, \mathrm{~V}_2 \mathrm{O}_4$ and $\mathrm{V}_2 \mathrm{O}_5$ upon reaction with alkali leads to formation of an oxide anion. The oxidation state of V in the oxide anion is:

1)  +3
 
2) +7
3) +5
4) +4

Correct answer: (3)

Solution:

$\mathrm{V}_2 \mathrm{O}_3$ - basic
$\mathrm{V}_2 \mathrm{O}_4$ - weakly acidic or amphoteric
$\mathrm{V}_2 \mathrm{O}_5$ - amphoteric

$
\mathrm{V}_2 \mathrm{O}_5+\text { alkali } \rightarrow \mathrm{VO}_4^{3-}
$

In the $\mathrm{VO}_4^{3-}$ ion, vanadium is in a +5 oxidation state.

Hence, the correct answer is option (3).

Question 22: Consider the following reactions: $\begin{aligned}
& \mathrm{A}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CrO}_2 \mathrm{Cl}_2+\text { Side Products } \\
& \quad \text { Little } \\
& \text { amount } \\
& \mathrm{CrO}_2 \mathrm{Cl}_{2(\text { Vapour })}+\mathrm{NaOH} \rightarrow \mathrm{~B}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O} \\
& \mathrm{~B}+\mathrm{H}^{+} \rightarrow \mathrm{C}+\mathrm{H}_2 \mathrm{O}
\end{aligned}$ The number of terminal ' O ' present in the compound ' C ' is______ .

1) 6
2) -
3) -
4) -

Correct answer: (1)

Solution:

$\mathrm{Cr}_2 \mathrm{O}_7^{2-}+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{CrO}_2 \mathrm{Cl}_2$

$\begin{aligned} & \mathrm{CrO}_2 \mathrm{Cl}_2(\text { Vapour })+\mathrm{NaOH} \rightarrow \\ & \mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{NaCl}+\mathrm{H}_2 \mathrm{O}\end{aligned}$

$
\mathrm{Na}_2 \mathrm{CrO}_4+\mathrm{H}^{\oplus} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{H}_2 \mathrm{O}
$

(C)

$\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7 \rightarrow 2 \mathrm{Na}^{\oplus}+\mathrm{Cr}_2 \mathrm{O}_7^{2-}$

No. of terminal " O " =6

Hence, the answer is 6.

Question 23: Consider the following reactions $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{KOH}}[\mathrm{~A}] \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{H}_2 \mathrm{SO}_4}[\mathrm{~B}]+\mathrm{K}_2 \mathrm{SO}_4$ The products $[\mathrm{A}]$ and $[\mathrm{B}]$, respectively are :

1) $\mathrm{K}_2 \mathrm{Cr}(\mathrm{OH})_6$ and $\mathrm{Cr}_2 \mathrm{O}_3$
 
2) $\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{Cr}_2 \mathrm{O}_3$
 
3) $\mathrm{K}_2 \mathrm{CrO}_4$ and $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$
 
4) K2CrO4 and CrO

Correct answer: (3)

Solution:

$\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{KO}} \mathrm{K}_2 \mathrm{CrO}_4 \xrightarrow[-\mathrm{H}_2 \mathrm{O}]{\mathrm{H}_2 \mathrm{SO}_4} \mathrm{~K}_2 \mathrm{Cr}_2 \mathrm{O}_7+\mathrm{K}_2 \mathrm{SO}_4$

Hence, the correct answer is option (3).

Question 24: The number of elements from the following that do not belong to lanthanoids is :
$\mathrm{Eu}, \mathrm{Cm}, \mathrm{Er}, \mathrm{Tb}, \mathrm{Yb}$ and $\mathrm{Lu}$

1) 3
2) 4
3) 1
4) 5

Correct answer: (3)

Solution:

Cm is an actinide

Hence, the answer is option (3).

Topic-Wise Weightage and Number of Questions

The table below shows the distribution of questions asked from the chapter D- and F- Block Elements in the January and April JEE Main 2026 exams. The table will help students get an idea about the topics important from the point of view of the exam and on which they should focus․

Last 10 Years' Weightage of D- and F- Block Elements in JEE Main

The last 10 years’ weightage of D- and F- Block Elements in JEE Main helps students understand the chapter’s importance and identify frequently asked topics. The table below highlights the topic-wise question distribution to help you prioritise your preparation and focus on high-weightage areas of the past ten years.

Also Read: Last 10 Years JEE Main Question Paper

Important Formulas of D- and F-Block Elements for JEE Main 2027

1. Magnetic Moment (Spin-Only Formula): $\mu=\sqrt{n(n+2)} \mathrm{BM}$

2. Number of Unpaired Electrons: $n=$ number of unpaired electrons

3. Oxidation State: Oxidation state = charge on ion (for a monoatomic ion)

4. Effective Atomic Number (EAN): $\mathrm{EAN}=Z-$ oxidation state + ligand electrons donated

5. Maximum Oxidation State of Transition Elements: Generally, it is related to the total number of $n s$ and $(n-1) d$ electrons.

6. Electronic Configuration of Transition Elements: $(n-1) d^{1-10} n s^{0-2}$

7. Electronic Configuration of Lanthanoids: $[X e] 4 f^{0-14} 5 d^{0-1} 6 s^2$

8. Electronic Configuration of Actinoids: $[R n] 5 f^{0-14} 6 d^{0-1} 7 s^2$

9. Magnetic Moment for $n$ Unpaired Electrons: $\mu=\sqrt{n(n+2)} \mathrm{BM}$

10. Lanthanide Contraction: Gradual decrease in atomic and ionic radii with increasing atomic number across the lanthanoid series.

Important Exceptions in D- and F-Block Elements for JEE Main 2027