JEE Chemistry - MCQ Practice Questions

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.