NEET 2013 Chemistry Question Paper with Answer and Solution

(D) Compound $(III)$ is the most stable because it is a conjugated enol and is further stabilized by intramolecular hydrogen bonding,which forms a stable six-membered ring.
Compound $(II)$ is the keto form,which is generally more stable than a non-conjugated enol.
Compound $(I)$ is a non-conjugated enol,making it the least stable.
Therefore,the stability order is $III > II > I$.

(D) The menstrual cycle is regulated by hormones. During the secretory phase,the corpus luteum secretes large amounts of progesterone,which is essential for maintaining the endometrium of the uterus. If the ovum is not fertilized,the corpus luteum degenerates. This degeneration leads to a sharp decrease in the levels of progesterone. The withdrawal of progesterone causes the breakdown of the endometrial lining and its blood vessels,which leads to menstruation.

(C) Facilitated transport is a type of passive transport where molecules move across the membrane along the concentration gradient with the help of special membrane proteins.
Key characteristics of facilitated transport include:
$1$. It is highly selective,allowing only specific molecules to pass through.
$2$. It shows transport saturation,meaning the rate of transport reaches a maximum when all protein transporters are occupied.
$3$. It does not require energy $(ATP)$,as it follows the concentration gradient (downhill transport).
$4$. It requires special membrane proteins.
'Uphill transport' refers to moving molecules against the concentration gradient,which is a characteristic of active transport,not facilitated transport.

(A) Biotin,also known as Vitamin $B_7$,is a water-soluble vitamin that acts as a coenzyme for carboxylase enzymes.
Chemically,biotin contains a ureido ring fused with a tetrahydrothiophene ring.
The presence of the tetrahydrothiophene ring means that sulphur $(S)$ is an essential structural component of the biotin molecule.
Therefore,sulphur is a constituent element of biotin.

(A) The nitrogen cycle involves several microbial processes that return nitrogen to the atmosphere.
$1$. Denitrification: This is the primary process where bacteria such as $Pseudomonas$ and $Thiobacillus$ reduce nitrate $(NO_3^-)$ to dinitrogen gas $(N_2)$.
$2$. Anaerobic Ammonium Oxidation (Anammox): This process involves the oxidation of ammonium $(NH_4^+)$ with nitrite $(NO_2^-)$ to produce dinitrogen gas $(N_2)$ under anaerobic conditions,primarily carried out by specialized bacteria like $Brocadia$ $anammoxidans$.
Both processes are essential for completing the nitrogen cycle by returning fixed nitrogen to the atmosphere.

(B) In aerobic respiration,the three major pathways are:
$1$. Glycolysis (Pathway $A$): Glucose is converted to pyruvate. Net products include $ATP$ and $NADH$.
$2$. Link Reaction/Krebs Cycle (Pathway $B$): Pyruvate is converted to Acetyl-$CoA$ and then enters the Krebs cycle. Net products include $ATP$ (or $GTP$),$NADH$,and $FADH_2$.
$3$. Electron Transport System (Pathway $C$): $NADH$ and $FADH_2$ are oxidized to produce $ATP$ and $H_2O$.
Looking at the diagram:
- Pathway $A$ (Glycolysis) produces $ATP$ (arrow $4$).
- Pathway $B$ (Krebs cycle) produces $ATP$ (arrow $8$).
- Pathway $C$ $(ETS)$ produces $ATP$ (arrow $12$).
Thus,arrows $4, 8,$ and $12$ represent the net production of $ATP$ in these three stages.

(D) Baked potatoes are primarily composed of starch,which is a complex carbohydrate.
$1$. In the mouth,the process of digestion begins with the action of salivary amylase (ptyalin) on starch,breaking it down into maltose.
$2$. As the food moves into the small intestine,pancreatic amylase continues the digestion of remaining starch into disaccharides.
$3$. Finally,in the small intestine,disaccharidases (such as maltase) present in the intestinal juice (succus entericus) break down disaccharides like maltose into glucose units for absorption.
Therefore,the correct sequence of enzymes acting on starch is salivary amylase $\rightarrow$ pancreatic amylase $\rightarrow$ disaccharidases.

(B) In a skeletal muscle fibre,the $A-$ band contains both actin and myosin filaments.
However,in the central part of the $A-$ band,there is a region where the thin actin filaments do not overlap with the thick myosin filaments.
This central region,which contains only thick myosin filaments,is known as the $H-$ zone.
Therefore,the $H-$ zone represents the central gap between the actin filaments that extend into the $A-$ band.

(D) In the given diagram of the human eye:
$A$ represents the Retina,which is the innermost layer of the eye and contains photoreceptor cells called rods and cones.
$B$ represents the Blind spot,an area where the optic nerve leaves the eye and photoreceptors are absent.
$C$ represents the Aqueous chamber,the space between the cornea and the lens,filled with aqueous humor.
$D$ represents the Sclera,the outermost layer of the eye.
Therefore,option $D$ is the correct identification and description.

(A) In $Spirogyra$, sexual reproduction occurs through conjugation where the gametes are morphologically similar (isogamous) but physiologically distinct. These gametes are non-flagellated (non-motile).
In $Volvox$ and $Chlamydomonas$, sexual reproduction can be isogamous, anisogamous, or oogamous, but the gametes are typically flagellated (motile).
In $Fucus$, the sexual reproduction is oogamous, involving large non-motile eggs and small motile sperms.

(C) The meristematic tissue responsible for the increase in the girth (secondary growth) of a tree trunk is the $Lateral \ meristem$ (also known as the $Cambium$ or $Lateral \ meristematic \ tissue$).
$Apical \ meristem$ is responsible for the increase in the length of the plant (primary growth).
$Intercalary \ meristem$ is found in the internodes and is responsible for the elongation of the stem in grasses and monocots.
$Phellem$ (cork) is a product of the cork cambium, which is a type of lateral meristem, but it is not the meristematic tissue itself.

(C) Coenzymes are small,non-protein organic molecules that bind to enzymes to facilitate their catalytic activity.
Many coenzymes are derived from vitamins,specifically water-soluble vitamins like $B$-complex vitamins.
For example,$NAD^+$ and $NADP^+$ contain niacin (vitamin $B_3$),while $FAD$ contains riboflavin (vitamin $B_2$).
Therefore,vitamins are the essential chemical components of many coenzymes.

(D) Primary productivity is defined as the rate at which biomass or organic matter is produced per unit area over a time period by plants during photosynthesis.
Secondary productivity is defined as the rate of formation of new organic matter by consumers.
Consumers (heterotrophs) ingest the organic matter produced by producers and convert it into their own biomass.
Therefore,the correct option is $D$.

(D) $CO$,$F^{-}$,and $PF_3$ possess lone pairs of electrons that they can donate to an electron-deficient species. Hence,they act as Lewis bases.
In contrast,$BF_3$ has an incomplete octet around the central Boron atom,which provides a vacant $p$-orbital.
Because of this vacant orbital,$BF_3$ acts as an electron-pair acceptor,making it a Lewis acid.
Therefore,$BF_3$ is the least likely to act as a Lewis base.

(D) Dipole-induced dipole interactions occur between a polar molecule and a non-polar molecule.
$HCl$ is a polar molecule because it has a permanent dipole moment $(\mu \neq 0)$.
$He$ atoms are non-polar because they are spherically symmetric and have no permanent dipole moment $(\mu = 0)$.
Therefore,the interaction between $HCl$ and $He$ is a dipole-induced dipole interaction.

(A) Heating $(NH_4)_2Cr_2O_7$ results in the evolution of nitrogen gas,water vapor,and chromium$(III)$ oxide: $(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + 4H_2O + N_2$.
$KClO_3$ and $Zn(ClO_3)_2$ are chlorates that decompose to release oxygen gas upon heating.
$K_2Cr_2O_7$ decomposes to release oxygen gas: $4K_2Cr_2O_7 \xrightarrow{\Delta} 4K_2CrO_4 + 2Cr_2O_3 + 3O_2$.

(C) Paramagnetism is the property of a substance that contains one or more unpaired electrons.
$1.$ $NO^{+}$: Total electrons = $7 + 8 - 1 = 14$. Molecular orbital configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2$. All electrons are paired,so it is diamagnetic.
$2.$ $CO$: Total electrons = $6 + 8 = 14$. Similar to $NO^{+}$,all electrons are paired,so it is diamagnetic.
$3.$ $O_2^-$: Total electrons = $8 + 8 + 1 = 17$. Molecular orbital configuration: $(\sigma 1s)^2, (\sigma^* 1s)^2, (\sigma 2s)^2, (\sigma^* 2s)^2, (\sigma 2p_z)^2, (\pi 2p_x)^2, (\pi 2p_y)^2, (\pi^* 2p_x)^2, (\pi^* 2p_y)^1$. It contains one unpaired electron in the $\pi^* 2p_y$ orbital,making it paramagnetic.
$4.$ $CN^{-}$: Total electrons = $6 + 7 + 1 = 14$. Similar to $CO$,all electrons are paired,so it is diamagnetic.

(A) For $NO$ (total $15$ electrons): Bond order $(B.O)$ $= 2.5$,magnetic behaviour is paramagnetic.
For $NO^{+}$ (total $14$ electrons): Bond order $(B.O)$ $= 3.0$,magnetic behaviour is diamagnetic.
Since the bond order increases,the bond energy also increases.
Thus,the process $NO \to NO^{+}$ satisfies both conditions.

(C) To determine the magnetic behavior,we calculate the total number of electrons in each species:
$1$. $NO^{+}$: $7 + 8 - 1 = 14$ electrons. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\sigma 2p_z)^2$. All electrons are paired,so it is diamagnetic.
$2$. $CO$: $6 + 8 = 14$ electrons. Similar to $NO^{+}$,all electrons are paired,so it is diamagnetic.
$3$. $O_2^{-}$: $8 + 8 + 1 = 17$ electrons. Configuration: $(\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x)^2 (\pi 2p_y)^2 (\pi^* 2p_x)^2 (\pi^* 2p_y)^1$. It has one unpaired electron,so it is paramagnetic.
$4$. $CN^{-}$: $6 + 7 + 1 = 14$ electrons. Similar to $CO$,all electrons are paired,so it is diamagnetic.
Therefore,$O_2^{-}$ is the only paramagnetic species.

(C) To determine the hybridization,we calculate the number of electron pairs around the central atom using the formula: $\text{Number of electron pairs} = \text{Number of bond pairs} + \text{Number of lone pairs}$.
For $NF_{3}$: The central atom $N$ has $3$ bond pairs and $1$ lone pair,totaling $4$ electron pairs,which corresponds to $sp^{3}$ hybridization.
For $H_{2}O$: The central atom $O$ has $2$ bond pairs and $2$ lone pairs,totaling $4$ electron pairs,which corresponds to $sp^{3}$ hybridization.
Since both $NF_{3}$ and $H_{2}O$ have $4$ electron pairs,both are $sp^{3}$ hybridized.

(B) Recombination frequency of $50 \%$ indicates that the genes are either located on different chromosomes or are so far apart on the same chromosome that they behave as if they are unlinked.
Genes that show $50 \%$ recombination frequency follow the law of independent assortment.
If genes are tightly linked,the recombination frequency is very low (much less than $50 \%$).
Therefore,the statement that 'the genes are tightly linked' is incorrect.

(D) During the $Zygotene$ stage of $Prophase-I$ in $Meiosis$,homologous chromosomes pair up,a process known as $Synapsis$.
Each pair of synapsed homologous chromosomes is referred to as a $Bivalent$ or a $Tetrad$ (since it consists of four chromatids).
$Axoneme$ is the core structure of cilia and flagella.
$Equatorial plate$ is the plane where chromosomes align during $Metaphase$.
$Kinetochore$ is the protein structure on chromatids where spindle fibers attach.