Out of Cu2Cl2 and CuCl2,
Question: Out of Cu2Cl2 and CuCl2, which is more stable and why? Solution: CuCl2 is more stable because Cu2+ has a higher electron density than Cu+. Cu2+ is smaller in size, has higher effective nuclear charge and therefore a higher hydration enthalpy ∆hyd of Cu2+, which makes it more stable....
Read More →When Cu2+ ion is treated with KI,
Question: When Cu2+ ion is treated with KI, a white precipitate is formed. Explain the reaction with the help of the chemical equation. Solution: 2Cu2++ 4I- Cu2I2+ I2 Cu2+gets reduced to Cu+, and Igets oxidized to I2....
Read More →Transition elements show high melting
Question: Transition elements show high melting points. Why? Solution: Transition elements have strong metallic bonds. Breaking those bonds becomes harder that means melting these elements will be difficult. So, melting points are higher for transition elements....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \mathrm{e}^{x}\left(\log x+\frac{1}{x^{2}}\right) d x$ Solution: Let $I=\int e^{x}\left(\log x+\frac{1}{x^{2}}\right) d x$ $=\int e^{x}\left(\log x+\frac{1}{x}-\frac{1}{x}+\frac{1}{x^{2}}\right) d x$ $=\int e^{x}\left(\log x-\frac{1}{x}\right) d x+\int e^{x}\left(\frac{1}{x}+\frac{1}{x^{2}}\right) d x$ Using integration by parts, $=e^{x}\left(\log x-\frac{1}{x}\right)-\int e^{x} \frac{d}{d x}\left(\log x-\frac{1}{x}\right) d x+\int e^{x}\left(\fr...
Read More →Why first ionisation enthalpy
Question: Why first ionisation enthalpy of Cr is lower than that of Zn? Solution: First ionisation enthalpy of Cr is lower than that of Zn. If we see the electronic configuration of Cr and Zn Cr [Ar] 3d5 4s1 Zn [Ar] 3d10 4s2 Removing an electron from a half-filled 4s orbital requires lesser energy than removing an electron from a filled stable 4s orbital....
Read More →Why E° values for Mn,
Question: Why E values for Mn, Ni and Zn are more negative than expected? Solution: Mn2+(3d5) and Zn2+(3d10) have half-filled and filled d orbitals which give them stability and therefore prefer to stay that way and not get reduced. As for Ni2+(3d8), it has very high negative hydration enthalpy which gets balanced by first and second ionization enthalpy....
Read More →Why does copper not replace
Question: Why does copper not replace hydrogen from acids? Solution: A positive reduction potential means the reduced form of Cu is more stable than hydrogen. Thus, Cu is less reactive than hydrogen and cannot displace it from acids....
Read More →Although +3 is the characteristic oxidation state
Question: Although +3 is the characteristic oxidation state for lanthanoids but cerium also shows +4 oxidation state because ___________. (i) it has variable ionisation enthalpy (ii) it tends to attain the noble gas configuration (iii) it tends to attain f 0 configuration (iv) it resembles Pb4+ Solution: Option (ii) and (iii) are the answers....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int e^{x}\left(\log x+\frac{1}{x}\right) d x$ Solution: Let $I=\int e^{x}\left(\log x+\frac{1}{x}\right) d x$ We know that $\int e^{x}\left\{f(x)+f^{\prime}(x)\right\}=e^{x} f(x)+c$ Here, $f(x)=\log x ; f^{\prime}(x)=\frac{1}{x}$ $\int e^{x}\left(\log x+\frac{1}{x}\right) d x=e^{x} \log x+c$...
Read More →Which of the following will not act
Question: Which of the following will not act as oxidising agents? (i) CrO3 (ii) MoO3 (iii) WO3 (iv) CrO42 Solution: Option (ii) and (iii) are the answers....
Read More →Transition elements form binary compounds
Question: Transition elements form binary compounds with halogens. Which of the following elements will form MF3 type compounds? (i) Cr (ii) Co (iii) Cu (iv) Ni Solution: Option (i) and (ii) are the answers....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{-x / 2} d x$ Solution: Let $\mathrm{I}=\int \frac{\sqrt{1-\sin \mathrm{x}}}{1+\cos \mathrm{x}} \mathrm{e}^{-\mathrm{x} / 2 \mathrm{dx}}$ put $\frac{x}{2}=t \Rightarrow x=2 t \Rightarrow d x=2 d t$ $\int \frac{\sqrt{1-\sin x}}{1+\cos x} e^{-x / 2} d x=2 \int \frac{\sqrt{1-\sin 2 t}}{1+\cos 2 t} e^{-t} d t$ $=2 \int \frac{\sqrt{\sin ^{2} t+\cos ^{2} t-2 \sin t \cos t}}{1+\cos 2 t} \mathrm{e}^{-t} \mathrm{dt}$ $=2...
Read More →Which of the following ions show higher spin
Question: Which of the following ions show higher spin only magnetic moment value? (i) Ti3+ (ii) Mn2+ (iii) Fe2+ (iv) Co3+ Solution: Option (ii) and (iii) is the answer....
Read More →Which of the following lanthanoids
Question: Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids? (i) Ce (ii) Eu (iii) Yb (iv) Ho Solution: Option (ii) and (iii) are the answers....
Read More →General electronic configuration of actinoids
Question: General electronic configuration of actinoids is (n2)f 114 (n-1)d02ns2. Which of the following actinoids have one electron in 6d orbital? (i) U (Atomic no. 92) (ii) Np (Atomic no.93) (iii) Pu (Atomic no. 94) (iv) Am (Atomic no. 95) Solution: Option (i) and (ii) are the answers....
Read More →Which of the following actinoids
Question: Which of the following actinoids show oxidation states up to +7? (i) Am (ii) Pu (iii) U (iv) Np Solution: Option (ii)Puand (iv) Np are the answers....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: Solution: Let I $=\int \frac{1+x}{(2+x)^{2}} e^{x} d x$ $=\int e^{x}\left\{\frac{(x+2)-1}{(x+2)^{2}}\right\} d x$ $=\int e^{x}\left\{\frac{1}{x+2}-\frac{1}{(x+2)^{2}}\right\}$ $=\int e^{x} \frac{1}{x+2} d x-\int e^{x} \frac{1}{(x+2)^{2}} d x$ Using integration by parts, $=\frac{e^{x}}{x+2}+\int e^{x} \frac{1}{(x+2)^{2}} d x-\int e^{x} \frac{1}{(x+2)^{2}} d x$ $=e^{x} \frac{1}{x+2}+c$...
Read More →In the form of dichromate,
Question: In the form of dichromate, Cr (VI) is a strong oxidising agent in acidic medium but Mo (VI) in MoO3 and W (VI) in WO3 are not because ___________. (i) Cr (VI) is more stable than Mo(VI) and W(VI). (ii) Mo(VI) and W(VI) are more stable than Cr(VI). (iii) Higher oxidation states of heavier members of group-6 of transition series are more stable. (iv) Lower oxidation states of heavier members of group-6 of transition series are more stable. Solution: Option (ii) and (iii) are the answers....
Read More →Transition elements show a magnetic moment
Question: Transition elements show a magnetic moment due to spin and orbital motion of electrons. Which of the following metallic ions have almost the same spin only magnetic moment? (i) Co2+ (ii) Cr2+ (iii) Mn2+ (iv) Cr3+ Solution: Option (i) and (iv) are the answers....
Read More →Generally, transition elements and their salts
Question: Generally, transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured? (i) KMnO4 (ii) Ce (SO4)2 (iii) TiCl4 (iv) Cu2Cl2 Solution: Option (i)KMnO4 and (ii)Ce (SO4)2 are the answers....
Read More →Why is HCl not used to make the medium
Question: Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in an acidic medium? (i) Both HCl and KMnO4 act as oxidising agents. (ii) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent. (iii) KMnO4 is a weaker oxidising agent than HCl. (iv) KMnO4 acts as a reducing agent in the presence of HCl. Solution: Option (ii)KMnO4 oxidises HCl into Cl2 which is also an oxidising agent is the answer....
Read More →Evaluate the following integrals:
Question: Evaluate the following integrals: $\int \frac{2-x}{(1-x)^{2}} e^{x} d x$ Solution: Let I $=\int \frac{2-x}{(1-x)^{2}} e^{x} d x$ $=\int e^{x}\left\{\frac{(1-x)+1}{(1-x)^{2}}\right\} d x$ $=\int e^{x}\left\{\frac{1}{1-x}+\frac{1}{(1-x)^{2}}\right\}$ $\frac{1}{1-x}=f(x) \frac{1}{(1-x)^{2}}=f^{\prime}(x)$ $=e^{x} \frac{1}{1-x}+c$...
Read More →Find the value
Question: Let $f: Z \rightarrow Z: f(x)=2 x$. Find $g: Z \rightarrow Z: g \circ f=l_{Z}$. Solution: To find: $g: Z \rightarrow Z: g \circ f=I_{Z}$ Formula used: (i) $f \circ g=f(g(x))$ (ii) g o f = g(f(x)) Given: (i) $g: Z \rightarrow Z:$ g of $=I_{Z}$ Solution: We have, $f(x)=2 x$ Let $f(x)=y$ ⇒ y = 2x $\Rightarrow \mathrm{x}=\frac{\mathrm{y}}{2}$ $\Rightarrow \mathrm{x}=\frac{\mathrm{y}}{2}$ Let $g(y)=\frac{y}{2}$ Where g: Z Z For g o f, ⇒ g(f(x)) ⇒ g(2x) $\Rightarrow \frac{2 x}{2}$ $\Rightarr...
Read More →Although Zirconium belongs to the 4d transition
Question: Although Zirconium belongs to the 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because___________. (i) both belong to d-block. (ii) both have the same number of electrons. (iii) both have a similar atomic radius. (iv) both belong to the same group of the periodic table Solution: Option (iii)both have a similar atomic radiusis the answer...
Read More →Highest oxidation state of manganese in fluoride is
Question: Highest oxidation state of manganese in fluoride is +4 (MnF4) but highest oxidation state in oxides is +7 (Mn2O7) because ____________. (i) fluorine is more electronegative than oxygen. (ii) fluorine does not possess d-orbitals. (iii) fluorine stabilises lower oxidation state. (iv) in covalent compounds, fluorine can form a single bond only while oxygen forms double bond. Solution: Option (iv)in covalent compounds, fluorine can form a single bond only while oxygen forms double bond. is...
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