Assertion (A): The carbonate of lithium decomposes easily
Question: Assertion (A): The carbonate of lithium decomposes easily on heating to form lithium oxide and CO2. Reason (R): Lithium being very small in size polarises large carbonate ion leading to the formation of more stable Li2O and CO2. (i) Both A and R are correct and R is the correct explanation of A. (ii) Both A and R are correct but R is not the correct explanation of A. (iii) Both A and R are not correct (iv) A is not correct but R is correct. Solution: Option (i)Both A and R are correct ...
Read More →Match the elements given in Column I
Question: Match the elements given in Column I with the colour they impart to the flame has given in Column II. Solution: (i) is f (ii) is d (iii) is b (iv) is c (v) is e (vi) is a...
Read More →Match the compounds given in Column I
Question: Match the compounds given in Column I with their uses mentioned in Column II. Solution: (i) is c (ii) is d (iii) is b (iv) is a...
Read More →Find the coordinates of the point on the curve
Question: Find the coordinates of the point on the curve $y^{2}=3-4 x$ where tangent is parallel to the line $2 x+y-2=0$. Solution: Given that the curve $y^{2}=3-4 x$ has a point where tangent is $\|$ to the line $2 x+y-2=0$. Slope of the given line is $-2$ $\because$ the point lies on the curve $\therefore y^{2}=3-4 x$ $\Rightarrow 2 y \frac{d y}{d x}=-4$ $\Rightarrow \frac{d y}{d x}=\frac{-2}{y}$ Now, the slope of the curve $=$ slope of the line $\Rightarrow \frac{-2}{y}=-2$ $\Rightarrow y=1$ ...
Read More →Match the elements given in Column I
Question: Match the elements given in Column I with the properties mentioned in Column II. Solution: (i) is c (ii) is b (iii) is d (iv) is a,e...
Read More →Find the equation of the perpendicular bisector of the line segment whose
Question: Find the equation of the perpendicular bisector of the line segment whose end points are A(10, 4) and B( - 4, 9). Solution: Perpendicular bisector: A perpendicular bisector is a line segment which is perpendicular to the given line segment and passes through its mid - point (or we can say bisects the line segment). Now to find the equation of perpendicular bisector first, we will find mid - point of the given line using mid - point formula (call it midpoint as M), $(\mathrm{x}, \mathrm...
Read More →What is the structure of BeCl2
Question: What is the structure of BeCl2 molecule in gaseous and solid-state? Solution: The gaseous/vapour state is different than the solid-state. The structure of BeCl2 in the solid-state is a polymeric chain structure. BeCl2 tends to form a chloro-bridged dimer at temperatures below 1200K and dissociates into a linear monomer at high temperatures of the order of 1200 K....
Read More →Why do beryllium and magnesium
Question: Why do beryllium and magnesium not impart colour to the flame in the flame test? Solution: Be and Mg electrons are tightly bound to the atom due to the small atomic and ionic size. The flame is due to the excitation of the electron from its energy states. The electrons of Be and Mgdo did not gain excitation from the energy provided by the flame. Hence they do not show any flame in the flame test....
Read More →In the Solvay process, can we obtain sodium carbonate
Question: In the Solvay process, can we obtain sodium carbonate directly by treating the solution containing (NH4) 2CO3 with sodium chloride? Explain. Solution: In the Solvay process, carbon dioxide is passed through a concentrated solution of sodium chloride saturated with ammonia, which forms ammonium carbonate followed by ammonium hydrogen carbonate. Ammonium hydrogen carbonate crystals separate and they are heated to form sodium carbonate. NH3 is recovered from the solution which contains NH...
Read More →Write the equation of the normal to the curve
Question: Write the equation of the normal to the curve $y=x+\sin x \cos x$ at $x=\frac{\pi}{2}$. Solution: Given that the curve $y=x+\sin x \cos x$ Differentiating both the sides w.r.t. $x$, $\frac{d y}{d x}=1+\cos ^{2} x-\sin ^{2} x$ Now, Slope of the tangent $\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{x}=\frac{\pi}{2}\right)=1+\cos ^{2} \frac{\pi}{2}-\sin ^{2} \frac{\pi}{2}$ $\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=1-1+0=0$ When $x=\frac{\pi}{2}, y=\frac{\pi}{2}$ Equation of the normal:...
Read More →All compounds of alkali metals are easily soluble in water
Question: All compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents. Explain. Solution: The alkali metal compounds form ionic compounds due to their large ionic size and low ionization enthalpy, while lithium forms compounds of covalent nature due to their small ionic size, high ionization enthalpy, and high electronegativity....
Read More →Why are BeSO4 and MgSO4 readily
Question: Why are BeSO4 and MgSO4 readily soluble in water while CaSO4, SrSO4 and BaSO4 are insoluble? Solution: BeSO4 and MgSO4 are readily soluble in water, CaSO4, SrSO4, and BaSO4 are insoluble. This is because the greater hydration enthalpies of Be2+ and Mg2+ ions overcome the lattice enthalpy factor and therefore their sulphate is soluble in water....
Read More →Discuss the trend of the following:
Question: Discuss the trend of the following: (i) Thermal stability of carbonates of Group 2 elements. (ii) The solubility and the nature of oxides of Group 2 elements. Solution: (i) The thermal stability of the carbonates increases with increasing cationic size. The more stable the oxide of an alkaline earth metal, the less stable is the carbonate of the same. Hence BeCO3 is highly unstable as BeO is stable. (ii) Alkali metals form oxides with oxygen and give metal oxides. The oxides will be ba...
Read More →Name an element from Group 2
Question: Name an element from Group 2 which forms an amphoteric oxide and a water-soluble sulphate. Solution: Beryllium oxide is amphoteric, unlike the other basic compounds....
Read More →Lithium resembles magnesium in some of its properties.
Question: Lithium resembles magnesium in some of its properties. Mention two such properties and give reasons for this resemblance. Solution: (i) Lithium and magnesium are both lighter and harder than the other metals in their respective groups. (ii) Halides of both elements, LiCl and MgCl2 are soluble in ethanol. These two elements have similar properties because of their similar atomic and ionic radii....
Read More →Complete the following reactions
Question: Complete the following reactions (i) O2-2 + H2O (ii) O-2 + H2O Solution: (i) O2-2 + 2H2O H2O2 + 2OH- (ii) O2+ 2H2O H2O2 + O2 + 2OH-...
Read More →When heated in air, the alkali metals form various
Question: When heated in air, the alkali metals form various oxides. Mention the oxides formed by Li, Na and K. Solution: Lithium forms monoxide, sodium forms peroxide, and potassium form superoxide....
Read More →How do you account for the strong reducing
Question: How do you account for the strong reducing power of lithium in aqueous solution? Solution: Lithium has the highest negative EÓ¨ value, which is 3.04V. Lithium has a small atomic size, the highest ionization enthalpy but it is compensated by its high hydration enthalpy. Due to this, the reducing power of lithium is highest in an aqueous solution....
Read More →Which of the following are the correct reasons
Question: Which of the following are the correct reasons for the anomalous behaviour of lithium? (i) The exceptionally small size of its atom (ii) Its high polarising power (iii) It has a high degree of hydration (iv) Exceptionally low ionisation enthalpy Solution: Option (i) and (ii) are the answers....
Read More →Choose the correct statements from the following.
Question: Choose the correct statements from the following. (i) Beryllium is not readily attacked by acids because of the presence of an oxide film on the surface of the metal. (ii) Beryllium sulphate is readily soluble in water as the greater hydration enthalpy of Be2+ overcomes the lattice enthalpy factor. (iii) Beryllium exhibits coordination number more than four. (iv) Beryllium oxide is purely acidic. Solution: Option (i) and (ii) are the answers....
Read More →Identify the correct formula of halides
Question: Identify the correct formula of halides of alkaline earth metals from the following. (i) BaCl2.2H2O (ii) BaCl2.4H2O (iii) CaCl2.6H2O (iv) SrCl2.4H2O Solution: Option (i)BaCl2.2H2O and (iii)CaCl2.6H2O are the answers....
Read More →When Zeolite, which is hydrated sodium aluminium
Question: When Zeolite, which is hydrated sodium aluminium silicate is treated with hard water, the sodium ions are exchanged with which of the following ion(s)? (i) H+ ions (ii) Mg2+ions (iii) Ca2+ions (iv) SO42-ions Solution: Option (ii)Mg2+ ions and (iii)Ca2+ ionsare the answers....
Read More →Write the angle made by the tangent
Question: Write the angle made by the tangent to the curve $x=e^{t} \cos t, y=e^{t} \sin t$ at $t=\frac{\pi}{4}$ with the $x$-axis. Solution: Given that the curve $x=e^{t} \cos t, y=e^{t} \sin t$ $\frac{d x}{d t}=e^{t} \cos t-e^{t} \sin t$ and $\frac{d y}{d t}=e^{t} \sin t+e^{t} \cos t$ $\Rightarrow \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{e^{t} \sin t+e^{t} \cos t}{e^{t} \cos t-e^{t} \sin t}=\frac{\sin t+\cos t}{\cos t-\sin t}$ Now, for $t=\frac{\pi}{4}$ $\frac{d y}{d x}=\f...
Read More →Which of the following compounds
Question: Which of the following compounds are readily soluble in water? (i) BeSO4 (ii) MgSO4 (iii) BaSO4 (iv) SrSO4 Solution: Option (i)BeSO4 and (ii)MgSO4 are the answers....
Read More →Write the coordinates of the point on the curve
Question: Write the coordinates of the point on the curve $y^{2}=x$ where the tangent line makes an angle $\frac{\pi}{4}$ with $x$-axis. Solution: Given that the curve $y^{2}=x$ has a point where the tangent line makes an angle $\frac{\pi}{4}$ with $x$-axis. $\therefore$ Slope of the tangent $\frac{\mathrm{dy}}{\mathrm{dx}}=\tan 45^{\circ}=1$ $\because$ the point lies on the curve. $y^{2}=x$ $\Rightarrow 2 y \frac{d y}{d x}=1$ $\Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$ $\Rightarrow \frac{1}{2 y...
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