AP EAMCET 2015 Chemistry Question Paper with Answer and Solution

(D) Ferric hydroxide sol $(Fe(OH)_3)$ is a positively charged sol. According to the Hardy-Schulze rule,the coagulating power of an electrolyte depends on the valency of the oppositely charged ion (anion in this case). The higher the valency of the flocculating ion,the greater is its coagulating power. The anions provided are $Cl^-$,$NO_3^-$,$SO_4^{2-}$,and $[Fe(CN)_6]^{3-}$. Since $[Fe(CN)_6]^{3-}$ has the highest negative charge (valency = $3$),it is the most effective in coagulating the positively charged ferric hydroxide sol.

(D) The number of moles of atoms is calculated as: $n = \frac{\text{Number of atoms}}{N_A} = \frac{3.011 \times 10^{22}}{6.022 \times 10^{23}} = 0.05 \ mol$.
The molar mass $(M)$ is given by: $M = \frac{\text{mass}}{n} = \frac{1.15 \ g}{0.05 \ mol} = 23 \ g/mol$.
Therefore,the atomic mass of the element is $23 \ u$.

(A) $1$. In $I_3^{-}$,the central $I$ atom has $3$ lone pairs and $2$ bond pairs,resulting in $sp^3d$ hybridisation and a linear shape.
$2$. In $BeCl_2$,the central $Be$ atom has $0$ lone pairs and $2$ bond pairs,resulting in $sp$ hybridisation and a linear shape.
$3$. Both species have a linear shape,but their hybridisation states ($sp^3d$ vs $sp$) are different.
$4$. Therefore,the pair $I_3^{-}$ and $BeCl_2$ satisfies the given condition.

(C) The nitrogen molecule $(N_2)$ has a triple bond $(N \equiv N)$.
Carbon monoxide $(CO)$ has a triple bond $(C \equiv O)$.
The acetylide ion $(C_2^{2-})$ has a triple bond $([C \equiv C]^{2-})$.
Nitric oxide $(NO)$ has $11$ valence electrons and a bond order of $2.5$,which corresponds to a bond between a double and a triple bond,but it does not possess a formal triple bond.

(A) The dipole moment $(\mu)$ of a molecule is a vector sum of the individual bond dipole moments.
In $1,4-$dichlorobenzene,the two $C-Cl$ bonds are oriented in opposite directions at an angle of $180^{\circ}$.
Since the bond dipoles are equal in magnitude and opposite in direction,they cancel each other out.
Therefore,the net dipole moment of $1,4-$dichlorobenzene is $\mu = 0$.

(B) According to the van't Hoff equation,the effect of temperature on the equilibrium constant $K_{eq}$ depends on the enthalpy change $\Delta H$ of the reaction.
For an exothermic reaction,$\Delta H < 0$,so increasing the temperature decreases the equilibrium constant $K_{eq}$.
For an endothermic reaction,$\Delta H > 0$,so increasing the temperature increases the equilibrium constant $K_{eq}$.

(C) $Fe(OH)_3$ sol is a positively charged sol. In electro-osmosis,the movement of the dispersion medium is observed under the influence of an electric field while the movement of sol particles is prevented. Since the sol particles are positively charged,they would naturally move towards the cathode. Therefore,in electro-osmosis,the dispersion medium moves in the opposite direction,i.e.,towards the anode.

(A) The correct matches are as follows:
$(A)$ Rubidium $(Rb)$ is an alkali metal,which is an $s$-block element. So,$A-3$.
$(B)$ Platinum $(Pt)$ has the atomic number $78$. So,$B-4$.
$(C)$ Eka-silicon was the name given by Mendeleev to Germanium $(Ge)$. So,$C-1$.
$(D)$ Polonium $(Po)$ is a radioactive element belonging to group $16$ (chalcogens). So,$D-2$.
Thus,the correct matching is $A-3, B-4, C-1, D-2$.

(D) . Ocean acts as a major sink for $CO_2$ through physical and biological processes. This statement is correct.
$B$. The greenhouse effect leads to the warming of the earth's surface,not cooling. This statement is incorrect.
$C$. Catalytic converters are used in automobiles to convert harmful $CO$ and unburnt hydrocarbons into less harmful $CO_2$ and $H_2O$. This statement is correct.
$D$. $H_2SO_4$ mist is formed by the oxidation of $SO_2$ in the atmosphere,but herbicides and insecticides are generally classified as particulate pollutants or pesticides,not specifically as mist. Therefore,$A$ and $C$ are the correct statements.

(B) Let $M$ be the mass of the earth and $m$ be the mass of the body. The escape velocity is given by $V_e = \sqrt{2GM/R}$.
The initial velocity of the body is $v = V_e/2 = \frac{1}{2} \sqrt{2GM/R}$.
Using the law of conservation of energy between the surface of the earth and the maximum height $h$:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$
Substituting $v^2 = \frac{1}{4} V_e^2 = \frac{1}{4} \left( \frac{2GM}{R} \right) = \frac{GM}{2R}$:
$\frac{1}{2}m \left( \frac{GM}{2R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$\frac{GMm}{4R} - \frac{GMm}{R} = - \frac{GMm}{R+h}$
$-\frac{3GMm}{4R} = - \frac{GMm}{R+h}$
$\frac{3}{4R} = \frac{1}{R+h}$
$3(R+h) = 4R$
$3R + 3h = 4R$
$3h = R$
$h = R/3$

(B) Baeyer's reagent is a cold,dilute,alkaline solution of $KMnO_4$.
It reacts with alkenes and alkynes to form vicinal diols,causing the characteristic purple color of $KMnO_4$ to disappear,which serves as a test for unsaturation.

(D) The basic character of a molecule depends on the availability of the lone pair of electrons on the central atom for donation.
In $PH_3$,the lone pair is present in an orbital with high $s$-character,making it less available for donation.
In $P(CH_3)_3$,the three methyl groups are electron-donating groups ($+I$ effect),which increase the electron density on the phosphorus atom,making the lone pair more available for donation.
Therefore,$P(CH_3)_3$ is a stronger base than $PH_3$.

(D) Temporary hardness of water is caused by the presence of magnesium and calcium bicarbonates,$Mg(HCO_3)_2$ and $Ca(HCO_3)_2$.
Clark's method involves the addition of calculated amounts of lime,$Ca(OH)_2$,to the water.
This process precipitates the bicarbonates as insoluble carbonates: $Ca(HCO_3)_2 + Ca(OH)_2 \rightarrow 2CaCO_3(s) + 2H_2O(l)$ and $Mg(HCO_3)_2 + 2Ca(OH)_2 \rightarrow 2CaCO_3(s) + Mg(OH)_2(s) + 2H_2O(l)$.
Therefore,Clark's method is specifically used for the removal of temporary hardness.

(D) The molecular mass of $NaOH = 23 + 16 + 1 = 40 \ g/mol$.
The number of moles of $NaOH = \frac{0.04 \ g}{40 \ g/mol} = 0.001 \ mol = 10^{-3} \ mol$.
The volume of the solution = $100 \ mL = 0.1 \ L$.
Molarity of $NaOH$ solution = $\frac{10^{-3} \ mol}{0.1 \ L} = 10^{-2} \ M$.
Since $NaOH$ is a strong base,$[OH^-] = 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Using the relation $pH + pOH = 14$,we get $pH = 14 - 2 = 12$.

(C) Given the quadratic equation: $\sqrt{2}x^2 - bx + (8 - 2\sqrt{5}) = 0$.
Let the roots of the equation be $\alpha$ and $\beta$.
From the properties of quadratic equations,the sum of the roots is $\alpha + \beta = \frac{-(-b)}{\sqrt{2}} = \frac{b}{\sqrt{2}}$.
The product of the roots is $\alpha\beta = \frac{8 - 2\sqrt{5}}{\sqrt{2}}$.
The harmonic mean $(HM)$ of two roots is given by $HM = \frac{2\alpha\beta}{\alpha + \beta}$.
Given $HM = 4$,we have:
$\frac{2\alpha\beta}{\alpha + \beta} = 4$
Substituting the values of $\alpha + \beta$ and $\alpha\beta$:
$\frac{2 \left( \frac{8 - 2\sqrt{5}}{\sqrt{2}} \right)}{\frac{b}{\sqrt{2}}} = 4$
$\frac{2(8 - 2\sqrt{5})}{b} = 4$
$2(8 - 2\sqrt{5}) = 4b$
$8 - 2\sqrt{5} = 2b$
$b = 4 - \sqrt{5}$.

(A) The inert pair effect refers to the reluctance of the $ns^2$ electron pair to participate in bonding due to poor shielding by $d$ and $f$ orbitals.
For lead $(Pb)$,which belongs to group $14$,the $+2$ oxidation state becomes more stable than the $+4$ oxidation state as we move down the group.
Therefore,lead forms compounds in the $+2$ oxidation state due to the inert pair effect.

(C) The acidic character of oxides increases as we move from left to right across a period in the periodic table due to an increase in electronegativity and oxidation state of the central atom.
Comparing the central atoms: $S$ (Group $16$),$P$ (Group $15$),$Cl$ (Group $17$),and $N$ (Group $15$).
Among these,$Cl$ is the most electronegative and exhibits the highest oxidation state $(+7)$ in $Cl_2O_7$.
Therefore,$Cl_2O_7$ is the most acidic oxide.
The order of acidic strength is: $SO_3 < P_4O_{10} < N_2O_5 < Cl_2O_7$.

(C) In deep-sea diving,$O_2$ is diluted with $He$ instead of $N_2$.
This is because $He$ has very low solubility in blood even under high pressure,which prevents the formation of bubbles in the blood (a condition known as bends) when the diver returns to the surface.

(A) Let the lines be $L_1: 2x + 3y - 1 = 0$,$L_2: x + 2y - 1 = 0$,and $L_3: ax + by - 1 = 0$.
Since the orthocentre is at the origin $(0, 0)$,the altitude from the vertex formed by $L_1$ and $L_2$ to $L_3$ must pass through the origin.
The intersection of $L_1$ and $L_2$ is found by solving $2x + 3y = 1$ and $x + 2y = 1$,which gives $x = -1$ and $y = 1$.
The altitude from $(-1, 1)$ to $L_3$ passes through $(0, 0)$,so its equation is $y = -x$,or $x + y = 0$.
Since this altitude is perpendicular to $L_3: ax + by - 1 = 0$,the slope of $L_3$ must be $1$. Thus,$a/b = -1$,or $a = -b$.
Substituting $b = -a$ into $L_3$,we get $ax - ay - 1 = 0$.
Since the orthocentre is the intersection of altitudes,the altitude from the intersection of $L_2$ and $L_3$ to $L_1$ must also pass through the origin.
The line perpendicular to $L_1$ passing through the origin is $3x - 2y = 0$.
Solving $3x - 2y = 0$ and $x + 2y = 1$ gives $x = 1/4$ and $y = 3/8$.
Substituting $(1/4, 3/8)$ into $ax - ay - 1 = 0$ gives $a(1/4 - 3/8) = 1$,so $a(-1/8) = 1$,which means $a = -8$.
Since $b = -a$,we have $b = 8$.
Thus,$(a, b) = (-8, 8)$.

(B) Let the required point be $P(x_1, y_1)$. Since $P$ lies on the line $4x - y - 2 = 0$,we have:
$4x_1 - y_1 - 2 = 0$ ... $(i)$
Given that $P$ is equidistant from $A(-5, 6)$ and $B(3, 2)$,we have $PA^2 = PB^2$:
$(x_1 + 5)^2 + (y_1 - 6)^2 = (x_1 - 3)^2 + (y_1 - 2)^2$
$x_1^2 + 10x_1 + 25 + y_1^2 - 12y_1 + 36 = x_1^2 - 6x_1 + 9 + y_1^2 - 4y_1 + 4$
$16x_1 - 8y_1 + 48 = 0$
Dividing by $8$,we get:
$2x_1 - y_1 + 6 = 0$ ... (ii)
Subtracting (ii) from $(i)$:
$(4x_1 - y_1 - 2) - (2x_1 - y_1 + 6) = 0$
$2x_1 - 8 = 0 \implies x_1 = 4$
Substituting $x_1 = 4$ in $(i)$:
$4(4) - y_1 - 2 = 0 \implies 16 - y_1 - 2 = 0 \implies y_1 = 14$
Thus,the point is $(4, 14)$.

(C) Given equation: $(x^2+y^2) \sin^2 \alpha = (x \cos \alpha - y \sin \alpha)^2$
Expanding the right side: $(x^2+y^2) \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$
Rearranging terms: $x^2 \sin^2 \alpha + y^2 \sin^2 \alpha = x^2 \cos^2 \alpha + y^2 \sin^2 \alpha - 2xy \sin \alpha \cos \alpha$
Simplifying: $x^2 \sin^2 \alpha = x^2 \cos^2 \alpha - 2xy \sin \alpha \cos \alpha$
$x^2(\sin^2 \alpha - \cos^2 \alpha) + 2xy \sin \alpha \cos \alpha = 0$
$-x^2 \cos(2\alpha) + xy \sin(2\alpha) = 0$
This is a homogeneous equation of the form $ax^2 + 2hxy + by^2 = 0$,where $a = -\cos(2\alpha)$,$h = \frac{1}{2} \sin(2\alpha)$,and $b = 0$.
The angle $\theta$ between the lines is given by $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$.
$\tan \theta = \left| \frac{2\sqrt{\frac{1}{4} \sin^2(2\alpha) - 0}}{-\cos(2\alpha) + 0} \right| = \left| \frac{\sin(2\alpha)}{-\cos(2\alpha)} \right| = |-\tan(2\alpha)| = |\tan(2\alpha)|$.
Thus,$\theta = 2\alpha$.

(C) Given that $A$ is a square matrix of order $n = 3$.
We know the property of the adjoint of a matrix: $|\operatorname{adj}(\operatorname{adj} M)| = |M|^{(n-1)^2}$.
Here,we need to find $|\operatorname{adj}(\operatorname{adj} A^2)|$.
Let $M = A^2$. Then,$|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^{(3-1)^2}$.
Since $(3-1)^2 = 2^2 = 4$,we have $|\operatorname{adj}(\operatorname{adj} A^2)| = |A^2|^4$.
Using the property $|A^k| = |A|^k$,we get $|A^2|^4 = (|A|^2)^4 = |A|^8$.
Therefore,the correct option is $|A|^8$.

(D) Alkali metals have only one valence electron per atom,which results in weak metallic bonding.
Due to these weak interatomic metallic bonds,alkali metals are soft and possess low melting and boiling points.
Therefore,both $(A)$ and $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) The magnitude of a vector $\overrightarrow{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $|\overrightarrow{a}| = \sqrt{x^2 + y^2 + z^2}$.
Calculating the magnitudes:
$m_1 = |\overrightarrow{a}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \approx 2.45$
$m_2 = |\overrightarrow{a}_2| = \sqrt{3^2 + (-4)^2 + (-4)^2} = \sqrt{9 + 16 + 16} = \sqrt{41} \approx 6.40$
$m_3 = |\overrightarrow{a}_3| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \approx 1.73$
$m_4 = |\overrightarrow{a}_4| = \sqrt{(-1)^2 + 3^2 + 1^2} = \sqrt{1 + 9 + 1} = \sqrt{11} \approx 3.32$
Comparing the values: $\sqrt{3} < \sqrt{6} < \sqrt{11} < \sqrt{41}$,which implies $m_3 < m_1 < m_4 < m_2$.

(A) The number of moles of the element is given by the ratio of the number of atoms to Avogadro's number $(N_A = 6.022 \times 10^{23} \ mol^{-1})$.
Number of moles $(n)$ = $\frac{3.011 \times 10^{22}}{6.022 \times 10^{23}} = 0.05 \ mol$.
Atomic mass $(M)$ = $\frac{\text{Mass}}{\text{Moles}} = \frac{1.15 \ g}{0.05 \ mol} = 23 \ g/mol$.
Therefore,the atomic mass of the element is $23$.

(C) To find the empirical formula,calculate the mole ratio of each element:
Carbon: $\frac{12.8}{12.0} = 1.066$
Hydrogen: $\frac{2.1}{1.008} = 2.083$
Bromine: $\frac{85.1}{79.9} = 1.065$
Dividing by the smallest value $(1.065)$:
Carbon: $\frac{1.066}{1.065} \approx 1$
Hydrogen: $\frac{2.083}{1.065} \approx 2$
Bromine: $\frac{1.065}{1.065} = 1$
Thus,the empirical formula is $CH_2Br$.
Empirical formula mass $= 12.0 + 2(1.008) + 79.9 = 93.996 \approx 94$.
Calculate the factor $n = \frac{\text{Molecular mass}}{\text{Empirical formula mass}} = \frac{187.9}{94} = 2$.
Molecular formula $= n \times (\text{Empirical formula}) = 2 \times (CH_2Br) = C_2H_4Br_2$.

(D) As we go higher in altitude,the atmospheric pressure decreases.
Since the boiling point of a liquid is the temperature at which its vapor pressure equals the external atmospheric pressure,a decrease in external pressure leads to a decrease in the boiling point.
Therefore,water boils at a lower temperature on the top of a mountain.

(B) Evaporation is a surface phenomenon.
When the surface area decreases,fewer molecules are exposed to the surface,leading to a decrease in the rate of evaporation.
Therefore,the statement that evaporation increases with a decrease in surface area is incorrect.

(D) Given that $3.011 \times 10^{22}$ atoms of an element weigh $1.15 \ g$.
We know that $1 \ mol$ of an element contains $N_A$ atoms,where $N_A \approx 6.022 \times 10^{23} \ atoms/mol$.
Mass of $6.022 \times 10^{23}$ atoms (molar mass) $= \frac{1.15 \ g}{3.011 \times 10^{22} \ atoms} \times 6.022 \times 10^{23} \ atoms/mol$.
$= 1.15 \times 20 = 23 \ g/mol$.
Therefore,the atomic mass of the element is $23 \ amu$.

(D) The de-Broglie wavelength $\lambda$ is related to kinetic energy $K.E.$ by the formula $\lambda = \frac{h}{\sqrt{2m(K.E.)}}$.
This implies $\lambda \propto \frac{1}{\sqrt{K.E.}}$
Given that the kinetic energy is reduced to half,$(K.E.)_2 = \frac{(K.E.)_1}{2}$.
Using the ratio $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{(K.E.)_1}{(K.E.)_2}} = \sqrt{\frac{(K.E.)_1}{(K.E.)_1 / 2}} = \sqrt{2}$.
Therefore,$\lambda_2 = \sqrt{2} \lambda_1$.

(C) For a $3d$-orbital,the principal quantum number $n=3$ and the azimuthal quantum number $l=2$.
For $l=2$,the magnetic quantum number $m$ can take any integer value from $-l$ to $+l$,which means $m \in \{-2, -1, 0, +1, +2\}$.
The spin quantum number $s$ can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.
Comparing these conditions with the given options,option $C$ $(n=3, l=2, m=-2, s=+\frac{1}{2})$ satisfies all the criteria.

(C) According to the first law of thermodynamics,$\Delta E = q + W$.
For an adiabatic process,there is no heat exchange with the surroundings,so $q = 0$.
Substituting $q = 0$ into the equation,we get $\Delta E = W$.
Since $W$ is the work done on the system,for expansion,the work done by the system is $W_{sys} = -W$.
Thus,$\Delta E = -W_{sys}$,which implies $\Delta W = -\Delta E$.

(D) Given equation is $(5+\sqrt{2}) x^2-b x+(8+2 \sqrt{5})=0$.
Let $\alpha$ and $\beta$ be the roots of this equation.
From the relation between roots and coefficients:
$\alpha+\beta = \frac{b}{5+\sqrt{2}}$
$\alpha \beta = \frac{8+2 \sqrt{5}}{5+\sqrt{2}}$
The harmonic mean $(HM)$ between the roots is given by $\frac{2 \alpha \beta}{\alpha+\beta} = 4$.
Substituting the values:
$\frac{2 \left( \frac{8+2 \sqrt{5}}{5+\sqrt{2}} \right)}{\frac{b}{5+\sqrt{2}}} = 4$
$\frac{2(8+2 \sqrt{5})}{b} = 4$
$\frac{8+2 \sqrt{5}}{b} = 2$
$b = \frac{8+2 \sqrt{5}}{2} = 4+\sqrt{5}$.

(B) The given reaction is the iodoform test for ethanol.
Ethanol $(C_2H_5OH)$ reacts with iodine $(I_2)$ in the presence of a base like sodium carbonate $(Na_2CO_3)$ to form iodoform $(CHI_3)$,which is also known as triiodomethane.
The balanced chemical equation is:
$C_2H_5OH + 4I_2 + 3Na_2CO_3 \longrightarrow CHI_3 + HCOONa + 5NaI + 3CO_2 + 2H_2O$
Therefore,'$X$' is $CHI_3$ (triiodomethane).

(A) When phenol is exposed to air,it undergoes slow oxidation to form $p$-benzoquinone (commonly referred to as quinone). The reaction is as follows:
$C_6H_5OH + [O] \rightarrow C_6H_4O_2$ (quinone)
This process is responsible for the pinkish color that phenol develops upon standing in air.

(A) The conversion of acetic acid $(CH_3COOH)$ to acetyl chloride $(CH_3COCl)$ is achieved using thionyl chloride $(SOCl_2)$. This is a standard reaction for converting carboxylic acids to acid chlorides.
The conversion of acetyl chloride $(CH_3COCl)$ to acetaldehyde $(CH_3CHO)$ is a Rosenmund reduction,which uses hydrogen gas in the presence of palladium catalyst poisoned with barium sulfate $(H_2 / Pd-BaSO_4)$.
Therefore,reagent $A$ is $SOCl_2$ and reagent $B$ is $H_2 / Pd-BaSO_4$.

(C) The given reaction sequence is the Sandmeyer reaction.
Step $1$: The conversion of primary aromatic amine $(Ar-NH_2)$ to benzene diazonium chloride $(Ar-\overset{+}{N} \equiv NCl^-)$ is called diazotization,which is carried out using $NaNO_2$ and $HCl$ at $0-5 \ ^\circ C$. Thus,$X = NaNO_2-HCl$.
Step $2$: The conversion of benzene diazonium chloride to aryl chloride $(Ar-Cl)$ is carried out using cuprous chloride $(CuCl)$ or copper powder in $HCl$. This is the Sandmeyer reaction. Thus,$Y = Cu/HCl$.

(A) The amino acid $Tryptophan$ contains an indole ring in its side chain.
An indole ring is a bicyclic structure consisting of a six-membered benzene ring fused to a five-membered nitrogen-containing pyrrole ring.
Among the given options,$Tryptophan$ is the only amino acid that possesses this specific structural feature.

(A) In methoxy methane $(CH_3-O-CH_3)$,the oxygen atom is $sp^3$ hybridized.
Due to the presence of two bulky $-CH_3$ groups,there is significant steric repulsion between them.
This repulsion causes the $C-O-C$ bond angle to increase from the ideal tetrahedral angle of $109^{\circ} 28^{\prime}$ to $111.7^{\circ}$.

(C) In electro-osmosis,the movement of the dispersed phase is prevented by a semi-permeable membrane,and the dispersion medium moves under the influence of an electric field.
Since $Fe(OH)_3$ sol is positively charged,the dispersion medium is negatively charged.
Therefore,the dispersion medium moves towards the anode.

(D) Analgesics are medicines used to relieve pain.
$L$-dopa is used for the treatment of Parkinson's disease.
Amoxycillin is an antibiotic.
Sulphapyridine is a sulpha drug used as an antibacterial agent.
Morphine is a potent narcotic analgesic used for the relief of severe pain,such as post-operative pain,cardiac pain,or pain associated with terminal cancer.

(C) To determine the oxidation state of the central metal ion,we equate the sum of oxidation states of all ligands and the metal to the overall charge of the complex.
For $Fe(CO)_5$:
Let the oxidation state of $Fe$ be $x$.
The ligand $CO$ (carbonyl) is a neutral ligand,so its oxidation state is $0$.
$x + (5 \times 0) = 0$
$x = 0$
Thus,the central metal ion $Fe$ in $Fe(CO)_5$ is in the zero oxidation state.

(B) The reaction of $AgNO_3$ with a coordination compound precipitates the chloride ions present outside the coordination sphere as $AgCl$.
Since $1 \ mole$ of $AgCl$ is formed from $1 \ mole$ of the complex,there must be only $1 \ Cl^-$ ion outside the coordination sphere.
The formula of the complex can be written as $[Co(NH_3)_X Cl_2]Cl$.
For $Co(III)$,the coordination number is $6$.
Therefore,$X + 2 = 6$,which gives $X = 4$.

(B) The cell reaction is: $Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \longrightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^{-}_{(aq)}$
The standard cell potential is calculated as: $E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 \ V - (-0.76 \ V) = 1.56 \ V$
The number of electrons transferred in the reaction is $n = 2$.
Using the formula $\Delta G^{\circ} = -nFE^{\circ}$:
$\Delta G^{\circ} = -2 \times 96500 \ C/mol \times 1.56 \ V$
$\Delta G^{\circ} = -301080 \ J/mol = -301.08 \ kJ/mol \approx -301 \ kJ/mol$

(A) The basicity of phosphines depends on the availability of the lone pair on the phosphorus atom.
In $P(CH_3)_3$,the three methyl groups exert a $+I$ (inductive) effect,which increases the electron density on the phosphorus atom.
This makes the lone pair more available for donation compared to $PH_3$,where no such electron-donating groups are present.
Therefore,the correct order of basic character is $P(CH_3)_3 > PH_3$.

(C) Misch metal is an alloy consisting of a lanthanide metal (about $95\%$),iron (about $5\%$),and traces of sulfur,carbon,calcium,and aluminum.
Therefore,the percentage of lanthanides and iron in misch metal is approximately $95\%$ and $5\%$,respectively.

(D) Levigation,also known as gravity separation,is a physical method used for the concentration of ores. It involves separating lighter gangue particles from heavier ore particles by washing them in a current of water. In contrast,$Calcination$,$Smelting$,and $Roasting$ are metallurgical processes that require heating the ore.

(C) Silicon is a group $14$ element. When it is doped with a group $15$ element like phosphorus,it has one extra electron compared to the silicon atoms. This excess electron makes it an $n$-type semiconductor.

(D) Assertion $(A)$ is true because noble gases are monoatomic and possess only weak van der Waals forces of attraction between their atoms,leading to very low boiling points.
Reason $(R)$ is also true because the general electronic configuration of noble gases (group $18$) is $ns^2 np^6$ (with $He$ being $1s^2$).
However,the electronic configuration is not the direct reason for the low boiling points; the weak intermolecular forces are. Therefore,$(R)$ is not the correct explanation of $(A)$.

(D) Bakelite is a thermosetting polymer formed by the condensation polymerization of phenol and formaldehyde in the presence of an acid or base catalyst.
During this process,the phenol molecules are linked together by methylene $(-CH_2-)$ bridges to form a cross-linked structure.

(C) According to Henry's law,the partial pressure of a gas is directly proportional to its mole fraction in the solution,given by $P = K_H \times \chi$.
Since the temperature remains constant,$K_H$ is constant.
When the pressure is doubled from $2 \ bar$ to $4 \ bar$,the mole fraction of the gas $(\chi_X)$ also doubles.
New mole fraction of gas $\chi_X' = 0.02 \times 2 = 0.04$.
The sum of mole fractions in a binary solution is $1$.
Therefore,the mole fraction of water $\chi_{H_2O} = 1 - \chi_X' = 1 - 0.04 = 0.96$.

(D) According to Raoult's law for a non-volatile solute,the relative lowering of vapour pressure is given by the mole fraction of the solute:
$\frac{p^{\circ} - p_s}{p^{\circ}} = \frac{n_2}{n_1 + n_2}$
Here,$n_2$ is the number of moles of glucose and $n_1$ is the number of moles of water.
$n_2 = \frac{18 \ g}{180 \ g/mol} = 0.1 \ mol$
$n_1 = \frac{90 \ g}{18 \ g/mol} = 5 \ mol$
For a dilute solution,$n_2$ in the denominator can be neglected:
$\frac{p^{\circ} - p_s}{p^{\circ}} \approx \frac{n_2}{n_1} = \frac{0.1}{5} = 0.02$

(D) Ferric hydroxide solution is a positively charged sol.
According to the Hardy-Schulze rule,the greater the valence of the coagulating ion,the greater is its power to cause coagulation.
$KCl \rightleftarrows K^{+} + Cl^{-}$
$KNO_3 \rightleftarrows K^{+} + NO_3^{-}$
$K_2SO_4 \rightleftarrows 2K^{+} + SO_4^{2-}$
$K_3[Fe(CN)_6] \rightleftarrows 3K^{+} + [Fe(CN)_6]^{3-}$
Since the ferric hydroxide sol is positively charged,it is coagulated by negative ions.
The valency of the negative ion is highest in $[Fe(CN)_6]^{3-}$ (i.e.,$3$).
Hence,$K_3[Fe(CN)_6]$ will be most effective in causing coagulation of ferric hydroxide solution.

(D) The basic character of a molecule depends on the availability of the lone pair of electrons on the central atom for donation.
In $PH_3$,the lone pair is present in an orbital with high $s$-character,making it less available for donation.
In $P(CH_3)_3$,the three methyl groups are electron-donating ($+I$ effect),which increases the electron density on the phosphorus atom,making the lone pair more available for donation.
Therefore,$P(CH_3)_3$ is a stronger base than $PH_3$.