JEE Chemistry - MCQ Practice Questions

Understanding:

We need to identify which amino acid among the options has a sulfur-containing side chain (R-group).

Step 1: Examine each option

Valine has the side chain $-\text{CH}({\text{CH}}_3{)}_2$, which contains only carbon and hydrogen.

Leucine has the side chain $-{\text{CH}}_2\text{CH}({\text{CH}}_3{)}_2$, also only C and H.

Threonine has the side chain $-\text{CH}(\text{OH}){\text{CH}}_3$, containing oxygen but no sulfur.

Cysteine has the side chain $-{\text{CH}}_2\text{SH}$, which contains a thiol ($-\text{SH}$) group with a sulfur atom.

Step 2: Recall sulfur-containing amino acids

The two common sulfur-containing amino acids are cysteine ($-{\text{CH}}_2\text{SH}$) and methionine ($-{\text{CH}}_2{\text{CH}}_2{\text{SCH}}_3$). Among the given options, only cysteine qualifies.

Answer:

Cysteine is the amino acid with a sulfur-containing side chain due to its thiol ($-\text{SH}$) group.

Quick Tip:

Cysteine's thiol group plays a critical role in forming disulfide bonds ($-\text{S}-\text{S}-$) between protein chains, which are essential for tertiary and quaternary protein structure.

Understanding:

We need to determine the orientation of the $-\text{OH}$ group at $\text{C}-1$ (anomeric carbon) in the Haworth projection of $\alpha$-$\text{D}$-glucose.

Step 1: Recall the convention for Haworth projections

In a Haworth projection of a $\text{D}$-sugar (pyranose ring), groups that are written on the right in the Fischer projection point downward (below the ring plane), and groups on the left point upward (above the ring plane).

Step 2: Define $\alpha$ and $\beta$ anomers

In the $\alpha$-anomer of a $\text{D}$-sugar, the $-\text{OH}$ at the anomeric carbon ($\text{C}-1$) is on the same side as the reference group (the $-{\text{CH}}_2\text{OH}$ at $\text{C}-5$ determines the $\text{D}$-configuration, which points upward). The $\alpha$ designation means the $-\text{OH}$ at $\text{C}-1$ is trans to the $-{\text{CH}}_2\text{OH}$ group, placing it below the ring plane.

Step 3: Confirm for $\alpha$-$\text{D}$-glucose

For $\alpha$-$\text{D}$-glucopyranose:

Answer:

In the Haworth projection of $\alpha$-$\text{D}$-glucose, the $-\text{OH}$ at $\text{C}-1$ is below the ring plane.

Quick Tip:

A simple rule: in $\alpha$-$\text{D}$-sugars, the $-\text{OH}$ at $\text{C}-1$ is down (below the ring); in $\beta$-$\text{D}$-sugars it is up (above the ring).

Understanding:

We need to identify the correct statement about the disaccharide sucrose.

Step 1: Analyse the structure of sucrose

Sucrose is a disaccharide formed by the condensation of $\alpha$-$\text{D}$-glucose and $\beta$-$\text{D}$-fructose. The glycosidic bond is formed between the anomeric carbon of glucose ($\text{C}-1$) and the anomeric carbon of fructose ($\text{C}-2$), giving an $\alpha$-$1,\beta$-$2$-glycosidic linkage, not a $\beta$-$1,4$ linkage.

Step 2: Check reducing sugar property

Because both anomeric carbons are involved in the glycosidic bond, there is no free anomeric $-\text{OH}$ group. Therefore, sucrose cannot open to give an aldehyde or ketone form, and it is a non-reducing sugar. It gives neither a positive Tollens' test nor a positive Fehling's test.

Step 3: Confirm the hydrolysis products

Hydrolysis of sucrose yields one molecule of $\text{D}$-glucose and one molecule of $\text{D}$-fructose. This is the correct statement.

Answer:

Sucrose hydrolyzes to give one molecule of glucose and one molecule of fructose.

Quick Tip:

The mixture of glucose and fructose obtained from sucrose hydrolysis is called invert sugar because the optical rotation changes from positive (sucrose, dextrorotatory) to negative (fructose dominates, levorotatory).

Understanding:

We need to identify which levels of protein structure are lost during denaturation.

Step 1: Define denaturation

Denaturation is the process by which a protein loses its native three-dimensional conformation due to disruption of non-covalent interactions (hydrogen bonds, hydrophobic interactions, electrostatic interactions) and disulfide bonds, without breaking the peptide bonds.

Step 2: Identify what is preserved

The primary structure — the sequence of amino acids linked by peptide bonds — is maintained during denaturation because peptide bonds are covalent bonds that are not broken by typical denaturing agents such as heat, urea, or $\text{pH}$ changes.

Step 3: Identify what is lost

All higher-order structures that depend on non-covalent interactions are disrupted:

All three are lost upon denaturation.

Answer:

Denaturation disrupts secondary, tertiary, and quaternary structures while the primary structure (peptide bond sequence) remains intact.

Quick Tip:

Renaturation (refolding) is possible in some cases if the denaturing agent is removed gently, showing that the primary sequence contains all the information needed to regain the native fold.

Understanding:

We need to identify which nitrogenous bases found in nucleic acids are classified as purines.

Step 1: Recall the classification of nitrogenous bases

Nitrogenous bases in nucleic acids are classified as:

Step 2: Assign each base to its class

Purines: Adenine ($\text{A}$) and Guanine ($\text{G}$) — both have a double-ring system.

Pyrimidines: Cytosine ($\text{C}$), Thymine ($\text{T}$, in DNA), and Uracil ($\text{U}$, in RNA) — all have a single ring.

Step 3: Match to the options

Among the choices, only Adenine and Guanine are purines.

Answer:

The purine bases in nucleic acids are Adenine and Guanine.

Quick Tip:

A useful mnemonic: "Pure As Gold" — Purines are Adenine and Guanine. The word 'purine' itself is longer (like the double ring), while 'pyrimidine' names three single-ring bases.

Understanding:

We need to identify the correct description of the peptide bond ($-\text{CO}-\text{NH}-$) that links amino acids in a protein chain.

Step 1: Recall the formation of the peptide bond

The peptide bond is a covalent amide bond formed between the carboxyl group ($-\text{COOH}$) of one amino acid and the amino group ($-{\text{NH}}_2$) of the next, with the loss of water:

Step 2: Understand the resonance structure

The lone pair on nitrogen delocalizes into the $\text{C}=\text{O}$ bond:

This resonance gives the peptide bond approximately 40% double bond character.

Step 3: Consequences of partial double bond character

Because of this partial double bond character:

Answer:

The peptide bond has partial double bond character due to resonance and is therefore planar and rigid.

Quick Tip:

In the trans configuration (most common), the two $\alpha$-carbons flanking the peptide bond are on opposite sides, minimising steric clashes between side chains.

Understanding:

We need to identify the monosaccharide products obtained when lactose is completely hydrolyzed.

Step 1: Recall the structure of lactose

Lactose is a disaccharide found in milk. It is formed by a $\beta$-$1,4$-glycosidic linkage between:

Step 2: Write the hydrolysis reaction

Step 3: Note the reducing nature

In lactose, the anomeric $-\text{OH}$ of glucose is free (only galactose's anomeric carbon is used in the bond), so lactose is a reducing sugar. Upon hydrolysis, one molecule each of galactose and glucose are produced.

Answer:

Complete hydrolysis of lactose gives one molecule of glucose and one molecule of galactose.

Quick Tip:

Lactose intolerance arises from deficiency of the enzyme lactase ($\beta$-galactosidase), which cleaves the $\beta$-$1,4$-glycosidic bond in lactose.

Understanding:

We need to classify the given proteins and identify which one is a fibrous protein.

Step 1: Recall the two major structural categories of proteins

Step 2: Classify each option

Step 3: Identify the answer

Only collagen is a fibrous protein among the given options.

Answer:

Collagen is a fibrous protein, characterized by its triple-helix structure and structural role in connective tissues.

Quick Tip:

Collagen is the most abundant protein in the human body. Its triple helix is stabilized by hydrogen bonds and requires hydroxyproline (formed by post-translational hydroxylation of proline, requiring Vitamin C). Deficiency of Vitamin C disrupts collagen synthesis, causing scurvy.

Understanding:

We need to correctly define the secondary structure of a protein and distinguish it from primary, tertiary, and quaternary structures.

Step 1: Define the four levels of protein structure

Step 2: Match to the options

Answer:

Secondary structure refers to the regular, repeating local conformations of the polypeptide backbone, such as the $\alpha$-helix and $\beta$-pleated sheet.

Quick Tip:

The key distinction: secondary structure involves only backbone atoms (no side chains), while tertiary structure involves side-chain ($\text{R}$-group) interactions.

Understanding:

We need to identify which monosaccharide is classified as both a ketone-containing sugar (ketose) and a six-carbon sugar (hexose) — i.e., a ketohexose.

Step 1: Define aldoses and ketoses

Step 2: Classify by carbon number

A hexose has 6 carbons.

Step 3: Classify each option

Step 4: Confirm fructose structure

The molecular formula of fructose is ${\text{C}}_6{\text{H}}_{12}{\text{O}}_6$, with the carbonyl group at $\text{C}-2$, making it a ketohexose.

Answer:

$\text{D}$-Fructose is the ketohexose among the given options, with a keto group at $\text{C}-2$.

Quick Tip:

Despite being a ketone, fructose is a reducing sugar because it can isomerise to an aldose form in alkaline conditions (via enolization), allowing it to reduce Tollens' and Fehling's reagents.

Understanding:

We need to find the number of geometrical isomers of $[\text{Co}(\text{en}{)}_2{\text{Cl}}_2{]}^{+}$, where en is ethylenediamine (a bidentate ligand).

Formula:

For an octahedral complex of the type $[\text{M}(\text{AA}{)}_2{\text{X}}_2{]}^{n+}$ where AA is a symmetric bidentate ligand, geometrical isomers arise from the relative positions of the two monodentate ligands $\text{X}$.

Step 1: Identify possible arrangements of the two ${\text{Cl}}^{-}$ ligands

The two ${\text{Cl}}^{-}$ ligands can be placed either:

Step 2: Check for optical isomers (not asked, but relevant)

The cis isomer is non-superimposable on its mirror image, so it is optically active (exists as a pair of enantiomers). The trans isomer has a plane of symmetry and is optically inactive. However, both cis and trans count as two distinct geometrical isomers.

Step 3: Count geometrical isomers

There are exactly 2 geometrical isomers: cis and trans.

Answer:

The complex $[\text{Co}(\text{en}{)}_2{\text{Cl}}_2{]}^{+}$ has 2 geometrical isomers (cis and trans).

Quick Tip:

For $[\text{M}(\text{AA}{)}_2{\text{X}}_2{]}^{n+}$, always remember: 2 geometrical isomers (cis & trans), but the cis form additionally shows optical isomerism giving a pair of enantiomers — don't confuse geometrical isomers with optical isomers.

Understanding:

We need to find the oxidation state of $\text{Fe}$ in ${\text{K}}_2[\text{Fe}(\text{CN}{)}_6]$.

Formula:

Let the oxidation state of $\text{Fe}$ be $x$. Then:

Step 1: Set up the equation

Step 2: Solve for $x$

Step 3: Verify the IUPAC name

${\text{K}}_2[\text{Fe}(\text{CN}{)}_6]$ is potassium hexacyanoferrate(II), confirming $\text{Fe}$ is in the $+2$ oxidation state. This is commonly known as potassium ferrocyanide.

Answer:

The oxidation state of iron in ${\text{K}}_2[\text{Fe}(\text{CN}{)}_6]$ is $+2$.

Quick Tip:

Distinguish: ${\text{K}}_2[\text{Fe}(\text{CN}{)}_6]$ → ${\text{Fe}}^{2+}$ (ferrocyanide); ${\text{K}}_3[\text{Fe}(\text{CN}{)}_6]$ → ${\text{Fe}}^{3+}$ (ferricyanide). A common exam trap is mixing up these two.

Understanding:

We need to state the relationship between the crystal field splitting energy in octahedral (${\Delta }_o$) and tetrahedral (${\Delta }_t$) complexes.

Formula:

From Crystal Field Theory, for the same metal ion and same ligands:

Step 1: Reasoning behind the relationship

Two factors contribute to the smaller splitting in tetrahedral complexes:

Step 2: Quantitative derivation basis

The ratio arises as:

Step 3: Consequence

Because ${\Delta }_t\approx 0.44{\Delta }_o<{\Delta }_o$, tetrahedral complexes almost always have a small crystal field splitting, making them usually high-spin regardless of the ligand.

Answer:

The crystal field splitting in a tetrahedral complex is $\frac{4}{9}$ of that in the corresponding octahedral complex.

Quick Tip:

Because ${\Delta }_t<P$ (pairing energy) in almost all cases, tetrahedral complexes are virtually always high-spin — a direct result of this $\frac{4}{9}$ factor.

Understanding:

We need to give the correct IUPAC name of the cis isomer of $[\text{Pt}({\text{NH}}_3{)}_2{\text{Cl}}_2]$.

Formula:

IUPAC naming rules for coordination compounds:

1. Name ligands alphabetically before the metal.

2. Anionic ligands end in '-o'; neutral ligands use common names (${\text{NH}}_3$ = ammine).

3. Prefixes (di, tri, etc.) are used for simple ligands.

4. Oxidation state of metal in parentheses.

5. Geometrical isomer prefix (cis/trans) is included.

Step 1: Identify oxidation state of Pt

Let oxidation state of Pt be $x$:

Step 2: Name the ligands alphabetically

Step 3: Assemble the full IUPAC name

Note: The correct spelling is 'ammine' (coordinated ${\text{NH}}_3$), NOT 'amine'. Option D is incorrect due to the spelling 'diamine'.

Answer:

The correct IUPAC name is cis-Diamminedichloroplatinum(II).

Quick Tip:

This complex is the anticancer drug cisplatin. Its trans isomer (transplatin) is pharmacologically inactive — a classic example of how geometrical isomerism affects biological activity.

Understanding:

We need to find the number of unpaired electrons in $[\text{Fe}(\text{CN}{)}_6{]}^{3-}$.

Formula:

The number of unpaired electrons depends on the crystal field splitting ${\Delta }_o$ vs. the pairing energy $P$:

Step 1: Find the oxidation state and electronic configuration of Fe

For $[\text{Fe}(\text{CN}{)}_6{]}^{3-}$:

So Fe is ${\text{Fe}}^{3+}$: ground state of Fe is $[\text{Ar}]3d^{6}4s^2$, so ${\text{Fe}}^{3+}$ is $[\text{Ar}]3d^5$.

Step 2: Apply Crystal Field Theory

${\text{CN}}^{-}$ is a strong field ligand (high in the spectrochemical series), so ${\Delta }_o>P$ → low-spin complex.

Step 3: Fill the $t_{2g}$ and $e_g$ orbitals for low-spin $d^5$

The $t_{2g}$ set has 3 orbitals holding 5 electrons: two orbitals are fully paired (4 electrons) and one orbital has 1 unpaired electron.

Number of unpaired electrons $=1$.

Answer:

The low-spin $d^5$ configuration $t_{2g}^5{e}_g^0$ gives exactly 1 unpaired electron.

Quick Tip:

Contrast with $[\text{Fe}({\text{H}}_2\text{O}{)}_6{]}^{3-}$: water is a weak field ligand, giving high-spin $d^5$ with 5 unpaired electrons. ${\text{CN}}^{-}$ forces pairing, reducing unpaired electrons from 5 to 1.

Understanding:

We need to identify which complex exhibits optical isomerism (i.e., exists as non-superimposable mirror images / enantiomers).

Formula:

A complex shows optical isomerism if it is chiral — it lacks a plane of symmetry, a centre of symmetry, and an improper rotation axis, making it non-superimposable on its mirror image.

Step 1: Analyse $[\text{Co}(\text{en}{)}_3{]}^{3+}$

This is an octahedral complex with three bidentate ethylenediamine ligands. It belongs to the $D_3$ point group and has no plane of symmetry. Its mirror image (the $\Lambda$ and $\Delta$ isomers) is non-superimposable. It is optically active.

Step 2: Analyse $\text{trans-}[\text{Co}(\text{en}{)}_2{\text{Cl}}_2{]}^{+}$

The trans isomer has a $C_2$ axis perpendicular to the $\text{Cl}$–$\text{Co}$–$\text{Cl}$ axis and a plane of symmetry containing both $\text{Cl}$ atoms and Co. It is optically inactive (achiral).

Step 3: Analyse $[\text{Ni}({\text{NH}}_3{)}_4{\text{Cl}}_2]$ (octahedral)

This complex of type $[{\text{MA}}_4{\text{B}}_2]$ has both cis and trans forms, but both possess planes of symmetry. No optical isomerism.

Step 4: Analyse $[{\text{CoCl}}_2(\text{en})({\text{NH}}_3{)}_2]$

This mixed-ligand complex in octahedral geometry possesses planes of symmetry in its principal isomeric forms. No optical isomerism.

Answer:

Only $[\text{Co}(\text{en}{)}_3{]}^{3+}$ is chiral and exists as $\Lambda$ and $\Delta$ optical isomers.

Quick Tip:

For tris-bidentate octahedral complexes like $[\text{M}(\text{AA}{)}_3{]}^{n+}$, optical isomerism is always present. Use the propeller analogy — a left-handed propeller ($\Lambda$) and right-handed ($\Delta$) are mirror images.

Understanding:

We need to interpret a magnetic moment of $5.92\text{BM}$ for $[{\text{MnBr}}_4{]}^{2-}$.

Formula:

The spin-only magnetic moment formula:

where $n$ = number of unpaired electrons.

Step 1: Back-calculate number of unpaired electrons

For $n=5$: $5\times 7=35$. So $n=5$ unpaired electrons.

Step 2: Confirm oxidation state of Mn

Let oxidation state of Mn be $x$:

${\text{Mn}}^{2+}$ has the configuration $[\text{Ar}]3d^5$ (5 $d$-electrons).

Step 3: Determine geometry and spin state

Answer:

The complex is high-spin tetrahedral with 5 unpaired electrons.

Quick Tip:

The spin-only values to memorise: $n=1\to 1.73$, $n=2\to 2.83$, $n=3\to 3.87$, $n=4\to 4.90$, $n=5\to 5.92\text{BM}$. A value near $5.92$ always means 5 unpaired electrons.

Understanding:

We need to identify the ambidentate ligand among the options.

Formula:

An ambidentate ligand is a ligand that can bond to the metal through two different donor atoms but bonds through only one at a time in a given complex.

Step 1: Evaluate each option

Step 2: Confirm the classic example

The nitrite ion ${\text{NO}}_2^{-}$ is the textbook example of an ambidentate ligand, as seen in linkage isomers:

Answer:

Nitrite (${\text{NO}}_2^{-}$) is the ambidentate ligand.

Quick Tip:

Ambidentate ligands give rise to linkage isomerism — a type of isomerism unique to coordination compounds. The other classic example is the thiocyanate ion ${\text{SCN}}^{-}$ (S-bonded or N-bonded).

Understanding:

We need to calculate the Effective Atomic Number (EAN) of $\text{Cr}$ in the complex $[\text{Cr}(\text{CO}{)}_6]$.

Formula:

The EAN rule:

For a neutral complex with a neutral metal ($\text{oxidation state}=0$):

Step 1: Determine oxidation state of Cr

CO is neutral; overall complex is neutral, so $\text{Cr}$ is in zero oxidation state.

Electrons on ${\text{Cr}}^0$: same as neutral Cr atom = 24 electrons.

Step 2: Count electrons donated by ligands

Each CO donates 2 electrons to the metal; with 6 CO ligands:

Step 3: Calculate EAN

36 is the atomic number of Krypton (Kr), confirming the 18-electron rule is satisfied.

Answer:

The EAN of Cr in $[\text{Cr}(\text{CO}{)}_6]$ is 36, equal to the nearest noble gas (Kr).

Quick Tip:

Metal carbonyls are the best examples of the 18-electron rule. $[\text{Cr}(\text{CO}{)}_6]$, $[\text{Fe}(\text{CO}{)}_5]$, and $[\text{Ni}(\text{CO}{)}_4]$ all obey it exactly with EAN = 36, 36, and 36 respectively.