NEET Practice Questions

Hint: \(P=\dfrac{V_{\text {total }}^2}{R_{\text {total }}}\)

Step: Find the power loss in the wire.
The total potential drop \((V_{total})\) is given by;
\(\Rightarrow V_{\text {total }}=\text { Potential drop per km } \times \text { Distance in km} \)
\(\Rightarrow V_{\text {total }}=8 ~\text{V}/ \text{km} \times 150 ~\text{km}=1200~\text{V}\)
The total resistance \((R_{total})\) over the distance is given by;
\(\Rightarrow R_{\text {total }}=\text { Resistance per km} \times \text { Distance in km} \)
\(\Rightarrow R_{\text {total }}=0.5~ \Omega / \text{km} \times 150 ~\text{km}=75 ~\Omega \)
The power loss \((P)\) in the wire is given by;
\(\Rightarrow P=\frac{V_{\text {total }}^2}{R_{\text {total }}}\)
\(\Rightarrow P=\frac{(1200 ~\text{V})^2}{75~\Omega} \Rightarrow P=19.2~\text{kW}\)
Hence, option (2) is the correct answer.

HINT: It is a square planar complex.

Explanation:

STEP 1: In [Ni(CN)₄]^2- 
Let oxidation state of Ni in [Ni(CN)₄]^2- = x. 
∴x – 4 = -2 
x= 2 
Ni²⁺ = [Ar] 3d⁸ 4s⁰ 
Electrons are filled in 3d according to Hund’s rule and Pauli’s exclusion principle. Since CN^– is a strong field ligand, it causes the pairing of all unpaired electrons.

STEP 2:

There are 2 unpaired electrons in d subshell which gets paired up. This leaves one empty 3d-orbital,1 4s orbital and 3 4p orbitals to accommodate 8 electrons by 4 CN^– ligands. 

Hence, hybridisation of [Ni(CN)₄]^2- is dsp².

HINT: SO₂ can decolorize acidified KMnO₄ solution.

Explanation: 

SO₂ gas can readily reduces acidified KMnO₄ solution because KMnO₄ is an oxidising agent that easily oxidise SO₂.

$2{\mathrm{MnO}}_4^{–}+$ $5{\mathrm{SO}}_2$ $+$ $2{\mathrm{H}}_2\mathrm{O}$ $\to 2{\mathrm{Mn}}^{2+}$ $+$ $5{\mathrm{SO}}_4^{2–}$ $+$ $4{\mathrm{H}}^{+}$

While other options such as NO₂ (strong oxidising agent), CO₂ (neither oxidising agent nor reducing agent cannot decolourise acidified KMnO₄ solution.

Hint: \(K.E_{\max }=h_f-\phi \)

Step: Find the velocity of the electron ejected from a silver surface by the ultraviolet light.
Given,
\( \lambda_0=3250 \times 10^{-10} \mathrm{~m}, ~~\lambda=2536 \times 10^{-10} \mathrm{~m} \)
\(\phi=\frac{{hc}}{\lambda_0}=\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{3250 \times 10^{-10}}=3.82 ~\mathrm{eV} \)
\({h} \nu=\frac{{hc}}{\lambda}=\frac{4.14 \times 10^{-15} \times 3 \times 10^8}{2536 \times 10^{-10}}=4.89~ \mathrm{eV}\)
According to Einstein’s photoelectric equation,
\(K.E_{\max }=h_f-\phi \)
\(K.E_{\max }=(4.89-3.82) \mathrm{eV}=1.077 \mathrm{eV}\)
\(\frac{1}{2} m v^2=1.077 \times 1.6 \times 10^{-19} \)
\(v=\sqrt{\frac{2 \times 1.077 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}} \)
\(\Rightarrow v=0.6 \times 10^6{\mathrm{~m}}/{\mathrm{~s}}\)
Hence, option (1) is the correct answer.

When an alkyl halide reacts with the base then an elimination reaction occurs and an alkene is obtained as a product. 
The reaction is as follows:

C₂H₅Br+ KOH(alc.)  → C₂H₄ + KBr+H₂O 

The mechanism of the reaction is as follows:

The other reactions are as follows:

CH₃CH₃+Cl₂ → CH₃CH₂Cl+ HCl ( Free radical substitution reaction)
CH₃Cl+ KOH(aq.)  → CH₃OH +KCl (nucleophilic substitution reaction)
CH₂=CH₂ +Br₂  → CH₂BrCH₂Br (electrophilic addition reaction)

1. Oxidising Reagent = number of oxygen atoms should be more

2. Peroxide or Superoxide bonds.

Fenton’s reagent is a solution of hydrogen peroxide (H₂O₂) with ferrous iron (typically iron(II) sulfate, FeSO₄) as a catalyst that is used to oxidize contaminants or waste waters.

 Osmium tetroxide (OsO4) is widely employed as a catalyst in various organic reactions.

The acidified potassium manganate(VII) solution oxidizes the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds

(a -> H₂) will evolve with Ph-CH₂-Br
(b -> AgCl) will form with Ph-CH₂-Br
(c -> Ph-CO₂H) will form with Ph-CH₂-Br

Hint: LiAlH₄ is a reducing agent

LiAlH₄ is a strong reducing agent. It will reduce aldehyde, ketone, and ester to alcohol.  LiAlH₄ is a hydride donor. The aldehyde converts into primary alcohol, and ketone converts into secondary alcohol, The ester is converted into primary alcohol.

The reaction is as follows:

Hint: Only primary amine gives carbylamine test

The carbylamine test is given by primary amine only. 

The structure of the given compound is as follows:

, , , 

The first two compounds are secondary amine, the third compound is a tertiary amine and the fourth compound is a primary amine.

Thus, only 2,4-dimethyl aniline gives the carbylamine test and the reaction is as follows: