IIT JEE 2025 Chemistry Question Paper with Answer and Solution in Hindi

$ NH _4 NO _2 \xrightarrow[60-70^{\circ} C ]{\Delta} N _2+2 H _2 O$
$NH _4 NO _3 \xrightarrow[200-250^{\circ} C ]{\Delta} N _2 O +2 H _2 O $

$\Delta_{O} \propto \frac{1}{\lambda}$
$\therefore$ The absorb wave length order is
$\left[Co(CN)_6\right]^{3-}<\left[Co\left(NH_3\right)_6\right]^{3+} < \left[Co\left(NH_3\right)_5 H_2 O\right]^{3+}<\left[Co\left(NH_3\right)_5 Cl\right]^{2+}$

$I ^{-}+2 MnO _4^{-}+ H _2 O \xrightarrow{\text { Neutral solution }} 2 MnO _2+ IO _3^{-}+2 OH ^{-}$

$(i) Ne _2 \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p_z^2\right)\left(\pi 2 p_{ x }^2=\pi 2 p_{ y }^2\right)\left(\pi^* 2 p_{ x }^2=\pi^* 2 p_{ y }^2\right)\left(\sigma^* 2 p_z^2\right)$
$B.O. =\frac{6-6}{2}=0$
$(ii) F _2 \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p _{ z }^2\right)\left(\pi 2 p _{ x }^2=\pi 2 p _{ y }^2\right)\left(\pi^* 2 p _{ x }^2=\pi^* 2 p _{ y }^2\right)$
$\text{HOMO}$
$(iii) O _2^{\oplus} \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p_z^2\right)\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y\right)$
$\text { B.O. }=\frac{6-1}{2}=2.5$
$O_2 \Rightarrow\left(\sigma 1 s^2\right)\left(\sigma^* 1 s^2\right)\left(\sigma 2 s^2\right)\left(\sigma^* 2 s^2\right)\left(\sigma 2 p_z^2\right)\left(\pi 2 p_x^2=\pi 2 p_y^2\right)\left(\pi^* 2 p_x^1=\pi^* 2 p_y^1\right)$
$B.O. \frac{6-2}{2}=2 ($Bond order increases, Bond strength increases$)$
$(iv)$ Size of atom increases, Bond length increases Size of $Li > B$
So, Bond length of $Li _2 >  B _2$

$La ^{+3} \rightarrow\left[{ }_{54} Xe \right] 4 f ^0$ diamagnetic
$Yb ^{+2} \rightarrow\left[{ }_{54} Xe \right] 4 f ^{14}$ diamagnetic
$Lu ^{+3} \rightarrow\left[{ }_{54} Xe \right] 4 f ^{14}$ diamagnetic
$La ^{+2} \rightarrow\left[{ }_{54} Xe \right] 5 d^1$ paramagnetic
$Ce ^{+4} \rightarrow\left[{ }_{54} Xe \right] 4 f ^0$ diamagnetic
$Ce ^{+3} \rightarrow\left[{ }_{54} Xe \right] 4 f ^1$ paramagnetic
$Yb ^{+3} \rightarrow\left[{ }_{54} Xe \right] 4 f ^{13}$ paramagnetic
$Lu ^{+2} \rightarrow\left[{ }_{54} Xe \right] 4 f ^{14} 5 d^1$ paramagnetic

$(A)$ Wrong: bond length order $Q > P > R$
$(B)$ Correct: Bond enthalpy $R > P > Q$
$(C)$ Wrong: $S _{ N } 2$ reactivity $Q > P > R$
$(D)$ Wrong: $pKa ( HI < HBr < HF )$ $R > P > Q ($ pKa $)$



 

For reduction of dichromate, balanced reaction is :
$Cr_2 O_7^{-2}(aq)+6 e^{-}+14 H^{+}(aq) \rightarrow 2 Cr^{3+}(aq)+7 H_2 O(l)$
$\qquad 3 mol \qquad 1 mol$
$\text { Number of Farads required }=3 mol$
$\text { Let current }=I \text { amperes }$
$\Rightarrow \frac{I \times 48.25 \times 60}{96500}=3$
$I=100 A$

Because concentration of $H +$ from weak acid is less we need to consider self ionization of $H _2 O$ also.
$HX ( aq ) \rightleftharpoons H ^{+}+ X ^{-}( aq )$
$10^{-3}- x \quad x + y \quad x$
$H _2 O ( l ) \rightleftharpoons H ^{+}( aq )+ OH ^{-}( aq )$
$ \quad x+y \quad y$
$\text { Approximation : }\left(10^{-3}-x\right) \simeq 10^{-3}$
$\Rightarrow \frac{x(x+y)}{10^{-3}}=K_{a}=4 \times 10^{-11}.....(1)$
$\Rightarrow y(x+y)=Kw=10^{-14}......(2)$
$\text { Add }(1)+(2)$
$\Rightarrow \quad(x+y)^2=5 \times 10^{-14}$
$\Rightarrow x+y=\left[H^{+}\right]=\sqrt{5} \times 10^{-7}$
$\Rightarrow x=\sqrt{5}=2.236$
Answer $2.23$ or $2.24 .$

$\left( P +\frac{ a }{ V _{ m }^2}\right)\left( V _{ m }- b \right)= RT$
​​​​​​$P V_m-b P+\frac{a}{V_m}-\frac{a b}{V_m^2}-R T=0$
$\Rightarrow P V_m^2-(b P+R T) V_m^2+a V_m-a b=0$
Coefficient of $V _{ m }^2=-( bP + RT )$
Coefficient of $V_m=a$
Ratio $=-\frac{(b P + RT )}{ a }=-\left[\frac{0.06 \times 300+24.6}{6}\right]=-7.1$

$H _2 O (l) \rightarrow H _2(g) + \frac{1}{2} O _2(g)$
$144 g \quad - \quad - $
$8 \ mol \ 8 \ mol - 8 \ mol$
$W=-P \Delta V=-(\Delta n) RT$
Change in gaseous moles $=12$
$\Rightarrow \quad W =-\frac{12 \times 8.3 \times 300}{1000} kJ=-29.88 kJ$.

Hexamethylene diamine series give $(+)$ carbylamine test.
$C _{ x } H _{ y } N _{ z }+(2 x +1 / 2) CuO \rightarrow xCO _2+\frac{ y }{2} H _2 O +\frac{ z }{2} N_2 ($Dumas method$)$
$C _6 H _{16} N_2$ then $\frac{ z }{2} N_2=\frac{2}{2} N_2=1$ mole
Given $10$ mole in this reaction so $w _{ N _2}=10 \times 28=280 gm$

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Weight of product of $S=350 \times 1 / 2=175 gm$ Ans.
* $\text{WES}$ : Williamson ether synthesis

$Mn ^{+2} \xrightarrow{ H _2 S+ NH _4 OH } \underset{\text { Pink/buff ppt. }}{ MnS \downarrow}$
$Ba ^{+2} \xrightarrow{\left( NH _4\right)_2 CO _3+ NH _4 OH } \underset{\text { White ppt. }}{ BaCO _3 \downarrow}$
$Al ^{+3} \xrightarrow{ NH _4 Cl + NH _4 OH } \underset{\text { White ppt. }}{ Al ( OH )_3 \downarrow}$
$Cu ^{+2} \xrightarrow{ H _2 S+ HCl \text { (dil.) }} \underset{\text { Black ppt. }}{ CuS \downarrow}$

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Gives $FeCl _3$ test because of presence of phenolic group and ninhydrin test.
image
Amino acid obtained gives $(+)$ Ninhydrin test but not phthalein dye test.
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Reaction with phenyl diazonium salt gives yellow dye.
image
Compound will form hydrazone derivative with aldehydic group of glucose

$Na _2\left[ Fe ( CN )_5( NO )\right]+ Na _2 S \rightarrow Na _4\left[ Fe ( CN )_5( NOS )\right]$
Sodium nitroprusside $\quad$ purple solution
 

$ICl \xrightarrow{ H _2 O } \underset{\substack{\downarrow \\ IO ^{\ominus}}}{ HIO }+ HCl$
$ClF _3 \xrightarrow{ H _2 O } \underset{\substack{\downarrow \\ CIO _2^{\ominus}}}{ HCIO _2}+3 HF$
$BrF _5 \xrightarrow{ H _2 O } \underset{\substack{\downarrow \\ BrO _3^{\ominus}}}{ HBrO _3+5 HF }$

$(i)$ Ion $-$ Ion $\rightarrow$ Interaction energy $\propto \frac{1}{r}$
Ion $-$ dipole $\rightarrow$ Interaction energy $\propto \frac{1}{ r ^2}$
Ion $-$ dipole Interaction energy approaches zero more rapidly as ' $r$ ' increases
$(ii)$ Rotating Polar molecules $\rightarrow$ Interaction energy $\propto \frac{1}{ r ^6}$.
$(iii)$ Dipole $-$ induced dipole forces are independent of temperature.
$(iv)$ Non$-$polar species show London dispersion forces.

$\text{CCP}:$
$(Z=4, a=400 pm, M=105.6 g / mol)$
$\text { Density }=\frac{Z \times M}{a^3 \times N_{A}}$
$=\frac{4 \times 105.6}{\left(400 \times 10^{-10}\right)^3 \times 6 \times 10^{23}}=11.00 g / \ cm^3$

$Ba \left( NO _3\right)_2( aq )+2 NaIO _3( aq ) \rightarrow Ba \left( IO _3\right)_2(s)+2 NaNO _3( aq )$
image
${\left[ NaIO _3\right]=\frac{6}{300}=2 \times 10^{-2} M }$
$Ba \left( IO _3\right)_2(s) \rightleftharpoons Ba ^{2+}+2 IO _3^{-}$
$\left(2 \times 10^{-2}+2 s\right)$
$K _{ sp }=\left[ Ba ^{+2}\right]\left[ IO _3^{-}\right]^2$
$1.58 \times 10^{-9}=\left[ Ba ^{+2}\right] \times\left(2 \times 10^{-2}\right)^2$
${\left[ Ba ^{+2}\right]= s =3.95 \times 10^{-6} M }$
$X =3.95$

$\frac{ x }{ m }= K \times C ^{1 / n }$
$\log \left(\frac{x}{m}\right)=\log K+\frac{1}{n} \log C$
$\log 4=\log K+\frac{1}{n} \log 10$
$0.6=\log K+\frac{1}{n}...(1)$
$\log 10=\log K+\frac{1}{n} \log 16$
$1=\log K+\frac{1}{n} \times 1.2...(2)$
Equation $(2) -$ equation $(1)$
$0.4=\frac{1}{n} \times(0.2) \Rightarrow n=0.5$
and $\log K =-1.4$
$\log \frac{x}{m}=\log K+\frac{1}{n} \times \log C$
$=-1.4+2 \times \log 20$
$=-1.4+2.6=1.2$
$\frac{x}{m}=10^{+1.2}=16$
$\left(\log 2=0.3,4 \log 2=1.2,16=10^{+1.2}\right)$
or
$\frac{x}{m}=K \times C^{1 / n}$
$4=K(10)^{1 / n}....(1)$
$10=K(16)^{1 / n}.....(2)$
$X=K(20)^{1 / n}.......(3)$
On solving equation $(1)$ and $(2)$
$\frac{1}{n}=2$
On solving equation $(1)$ and $(3)$
$\frac{4}{X}=\left(\frac{10}{20}\right)^2$
$X=16$
On solving equation $(2)$ and $(3)$
$\frac{10}{X}=\left(\frac{16}{20}\right)^2$
$X=15.625$

 $ A + R \rightarrow \text { Product }$
$t =0 A_0 10 A_0$
$t = t 0.6 A_0 9.6 A_0$
$\text { Rate }= k [ A ][ R ]$
$\text { Rate }_1= k \left(0.6 A_0\right) \times 9.6 A_0$
$A + R \rightarrow$ Product
$ t =0 A_0 10 A_0 \text { (excess) }$
$t = t 0.6 A_0 10 A_0$
$\text { Rate }= k ^{\prime}[ A ], k ^{\prime}= k [ R ]$
$\text { Rate }_2=\left( k \times 10 A_0\right) \times\left(0.6 A_0\right)$
$100 \times \frac{\Delta \text { Rate }}{\text { Rate }_1}=\frac{(0.6 \times 10-0.6 \times 9.6}{0.6 \times 9.6} \times 100=4.1666$

$\pi=\rho g h=10^3 \times 10 \times 2 \times 10^{-2} \text { Pascal }=200 \text { Pascal }$
$\pi=C R T$
$200=\frac{2}{M} \times 1000 \times 8.3 \times 300$
$M=24900=2.49 \times 10^4 g / mol$
$X=2.49$

$C _4 H _{10}(g)+\frac{13}{2} O _2(g) \rightarrow 4 CO _2(g)+5 H _2 O (l)$
$\Delta_{ r } G ^{ o }=4 \Delta_{ f } G _{ CO _2}^{ o }+5 \Delta_{ f } G _{ H _2 O }^{ o }-\Delta_{ f } G _{ C _4 H _{10}}^{ o }$
$=4 \times(-394)+5(-237)+18$
$=-2743 kJ / mol $
$\Delta_{ r } G ^{\circ}=- nFE ^{\circ}$
$-2743 \times 1000=-26 \times FE ^{\circ}$
$E ^{ o }=\frac{105.5}{F} \times 10^3=105.50$

Sum of spin magnetic moments of both complexes is $7.70$ to $7.73 B . M$.

$\underset{\text { M.M. }=1024}{\text { Octasaccharide }}+\underset{\text { M.M. }=126}{7 H _2 O } \longrightarrow \begin{array}{l}\text { Ribose }+2 \text { deoxyribose }+ \text { glucose } \\ \text { Totalmass }=1024+126=1150\end{array}$
$58.26=\frac{134 \times n}{1150} \times 100$
$\frac{66.999}{100}=134 n n=4.99=5$
$5$ units of $2-$Deoxyribose
$1150=(5 \times 150)+(x \times 150)+(y \times 180)$
$1150=\underbrace{750}_{5 \text { unit }}+\underbrace{1500}_{300 \text { unit }}+\underbrace{180 y }_{180 \text { unit }}$
$n =2.00$