Solve the following
Question: FeSO4solution mixed with (NH4)2SO4solution in 1:1 molar ratio gives the test of Fe2+ion but CuSO4solution mixed with aqueous ammonia in 1:4 molar ratio does not give the test of Cu2+ion. Explain why? Solution: Both the compounds i.e., $\mathrm{FeSO}_{4} \cdot\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4} \cdot 6 \mathrm{H}_{2} \mathrm{O}$ and $\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}\right] \mathrm{SO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}$ fall under the category of additio...
Read More →The diameters of internal and external surfaces of a hollow spherical
Question: The diameters of internal and external surfaces of a hollow spherical shell are $10 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively. If it is melted and recast into a solid cylinder of length of $2 \frac{2}{3} \mathrm{~cm}$, find the diameter of the cylinder. Solution: The internal and external radii of the hollow sphere are 3cm and 5cm respectively. Therefore, the volume of the spherical shell is $V=\frac{4}{3} \pi \times\left\{(5)^{3}-(3)^{3}\right\}$ $=\frac{4}{3} \pi \times 98 \mat...
Read More →Explain the bonding in coordination compounds in terms of Werner’s postulates.
Question: Explain the bonding in coordination compounds in terms of Werners postulates. Solution: Werners postulates explain the bonding in coordination compounds as follows: (i)A metal exhibits two types of valencies namely, primary and secondary valencies. Primary valencies are satisfied by negative ions while secondary valencies are satisfied by both negative and neutral ions. (In modern terminology, the primary valency corresponds to the oxidation number of the metal ion, whereas the seconda...
Read More →Calculate the overall complex dissociation equilibrium constant for
Question: Calculate the overall complex dissociation equilibrium constant for the Cu(NH3)42+ion, given that 4for this complex is 2.1 1013. Solution: 4= 2.1 1013 The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, 4. $\therefore \frac{1}{\beta_{4}}=\frac{1}{2.1 \times 10^{13}}$ $=4.7 \times 10^{-14}$...
Read More →The corresponding sides of two similar triangles are in the ratio of 2 : 3.
Question: The corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2, find the area of larger triangle. Solution: If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides. $\therefore \frac{\text { area of smaller triangle }}{\text { area of larger triangle }}=\left(\frac{\text { Side of smaller triangle }}{\text { Side of larger triangle }}\right)^{2}$ $\Rightarrow...
Read More →The hexaquo manganese(II) ion contains five unpaired electrons,
Question: The hexaquo manganese(II) ion contains five unpaired electrons, while the hexacyanoion contains only one unpaired electron. Explain using Crystal Field Theory. Solution: Hence, hexaaquo manganese (II) ion has five unpaired electrons, while hexacyano ion has only one unpaired electron....
Read More →Predict the number of unpaired electrons in the square planar
Question: Predict the number of unpaired electrons in the square planar [Pt(CN)4]2ion. Solution: In this complex, Pt is in the +2 state. It forms a square planar structure. This means that it undergoes dsp2hybridization. Now, the electronic configuration of Pd(+2) is 5d8. $\mathrm{CN}^{-}$being a strong field ligand causes the pairing of unpaired electrons. Hence, there are no unpaired electrons in $\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}$....
Read More →In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point
Question: In a trapezium ABCD, it is given that AB∥CD and AB = 2 CD. Its diagonals AC and BD intersect at a point O such that ar(△AOB) = 84 cm2Find ar(△COD) Solution: In △AOB and CODABO = CDO (Alternte angles in AB∥CD)AOB = COD (Vertically opposite angles)By AA similarity criterion, △AOB △CODIf two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides. $\therefore \frac{\operatorname{area}(\triangle \mathrm{AOB})}{\operatorname{are...
Read More →Solve the following
Question: Explain [Co(NH3)6]3+is an inner orbital complex whereas [Ni(NH3)6]2+is an outer orbital complex. Solution:...
Read More →△ABC ∼ △DEF such that ar(△ABC) = 64 cm2 and ar(△DEF)
Question: △ABC △DEF such that ar(△ABC) = 64 cm2and ar(△DEF) = 169 cm2If BC = 4 cm, find EF Solution: We have △ABC △DEFIf two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides. $\therefore \frac{\operatorname{area}(\triangle \mathrm{ABC})}{\operatorname{area}(\triangle \mathrm{DEF})}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2}$ $\Rightarrow \frac{64}{169}=\left(\frac{\mathrm{BC}}{\mathrm{EF}}\right)^{2}$ $\Rightarrow\left(\...
Read More →Solve the following
Question: [Fe(H2O)6]3+is strongly paramagnetic whereas [Fe(CN)6]3is weakly paramagnetic. Explain. Solution: In both $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ and $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}, \mathrm{Fe}$ exists in the $+3$ oxidation state i.e., in $d^{5}$ configuration. Since CNis a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only one unpaired electron left in thed-orbital. Therefore, $\mu=\sqrt{n(n+2)}...
Read More →Find the length of the altitude of an equilateral triangle of side 2a cm.
Question: Find the length of the altitude of an equilateral triangle of side 2acm. Solution: We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.Suppose AD is the altitude drawn from the vertex A to the side BC.So, It will bisects the side BCDC =aNow, In right triangle ADCBy using Pythagoras theorem, we have $\mathrm{AC}^{2}=\mathrm{CD}^{2...
Read More →Solve the following
Question: [NiCl4]2is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why? Solution: Though both [NiCl4]2and [Ni(CO)4] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Clis a weak field ligand and it does not cause the pairing of unpaired 3delectrons. Hence, [NiCl4]2is paramagnetic. In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d84s2. But CO is a strong field ligand. Therefore,...
Read More →A copper rod of diameter 1 cm and length
Question: A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. Solution: The radius of the copper rod is 0.5 cm and length is 8 cm. Therefore, the volume of the copper rod is $V=\pi \times(0.5)^{2} \times 8 \mathrm{~cm}^{3}$ Let the radius of the wire isrcm. The length of the wire is 18 m=1800 cm. Therefore, the volume of the wire is $V_{1}=\pi \times(r)^{2} \times 1800 \mathrm{~cm}^{3}$ Since, the volume of the c...
Read More →Explain on the basis of valence bond theory that
Question: Explain on the basis of valence bond theory that [Ni(CN)4]2ion with square planar structure is diamagnetic and the [NiCl4]2ion with tetrahedral geometry is paramagnetic. Solution: Ni is in the +2 oxidation state i.e., in d8configuration. There are 4 CNions. Thus, it can either have a tetrahedral geometry or square planar geometry. Since CNion is a strong field ligand, it causes the pairing of unpaired 3delectrons. It now undergoes dsp2hybridization. Since all electrons are paired, it i...
Read More →Give evidence that
Question: Give evidence that [Co(NH3)5Cl]SO4and [Co(NH3)5SO4]Cl are ionization isomers. Solution: When ionization isomers are dissolved in water, they ionize to give different ions. These ions then react differently with different reagents to give different products....
Read More →A vessel in the shape of a cuboid contains some water.
Question: A vessel in the shape of a cuboid contains some water. If three identical spheres immersed in the water, the level of water is increased by 2 cm. If the area of the base of the cuboid is 160 cm2and its height 12 cm, determine the radius of any of the spheres. Solution: The area of the base of the cuboid is $160 \mathrm{~cm}^{2}$. After immersing three identical spheres the level of the water is increased by $2 \mathrm{~cm}$. Therefore, the volume of the increased water is $V=160 \times...
Read More →Indicate the types of isomerism exhibited by the following complexes anddraw the structures for these isomers:
Question: Indicate the types of isomerism exhibited by the following complexes anddraw the structures for these isomers: K[Cr(H2O)2(C2O4)2 [Co(en)3]Cl3 [Co(NH3)5(NO2)](NO3)2 [Pt(NH3)(H2O)Cl2] Solution: i. Both geometrical (cis-, trans-) isomers for $\mathrm{K}\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}\right]$ can exist. Also, optical isomers for cis-isomer exist. Trans-isomer is optically inactive. On the other hand,cis-isomer is opt...
Read More →A ladder 10 m long reaches the window of a house 8 m above the ground.
Question: A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall is 6 m. Solution: Let AB be a ladder and B is the window at 8 m above the ground C.Now, In right triangle ABCBy using Pythagoras theorem, we have $\mathrm{AB}^{2}=\mathrm{BC}^{2}+\mathrm{CA}^{2}$ $\Rightarrow 10^{2}=8^{2}+\mathrm{CA}^{2}$ $\Rightarrow \mathrm{CA}^{2}=100-64$ $\Rightarrow \mathrm{CA}^{2}=36$ $\Rightarrow \mathrm{CA}=6 \mathrm{~m}$ ...
Read More →In the given figure, DE∥BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm.
Question: In the given figure, DE∥BC such that AD =x cm, DB = (3x+ 4) cm, AE = (x+ 3) cm and EC = (3x+ 19) cm. Find the value of x Solution: In △ADE and △ABCADE = ABC (Corresponding angles in DE∥BC)AED = ACB (Corresponding angles in DE∥BC)By AA similarity criterion, △ADE △ABCIf two triangles are similar, then the ratio of their corresponding sides are proportional $\therefore \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}$ $\Rightarrow \frac{\mathrm{AD}}{\mathrm{AD}+\mathrm{DB}}...
Read More →A copper sphere of radius 3 cm is melted and recast into a right
Question: A copper sphere of radius 3 cm is melted and recast into a right circular cone of height 3 cm. Find the radius of the base of the cone. Solution: The radius of the copper sphere is 3cm. Therefore, the volume of the copper sphere is $V=\frac{4}{3} \pi \times(3)^{3} \mathrm{~cm}^{3}$ The copper sphere is melted to produce a right circular cone. The height of the right circular cone is 3cm. Let the base-radius of the right circular cone isr. Then, the volume of the right circular cone is ...
Read More →Write the IUPAC names of the following coordination compounds:
Question: Write the IUPAC names of the following coordination compounds: (i)[Co(NH3)6]Cl3 (ii)[Co(NH3)5Cl]Cl2 (iii)K3[Fe(CN)6] (iv)K3[Fe(C2O4)3] (v)K2[PdCl4] (vi)[Pt(NH3)2Cl(NH2CH3)]Cl Solution: (i)Hexaamminecobalt(III) chloride (ii)Pentaamminechloridocobalt(III) chloride (iii)Potassium hexacyanoferrate(III) (iv)Potassium trioxalatoferrate(III) (v)Potassium tetrachloridopalladate(II) (vi)Diamminechlorido(methylamine)platinum(II) chloride...
Read More →The largest sphere is to be curved out of a right circular
Question: The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere. Solution: The radius of the right circular cylinder is 7cm and the height is 14cm. Therefore, the radius of the largest sphere curved out from the cylinder is the minimum of the radius and half the height of the cylinder, which is 7cm. Therefore, the volume of the sphere is $V=\frac{4}{3} \pi \times(7)^{3}$ $=1437.33 \mathrm{~cm}^{3}$...
Read More →If △ABC ∼ △DEF such that 2AB = DE and BC = 6 cm, find EF
Question: If △ABC △DEF such that 2AB = DE and BC = 6 cm, find EF Solution: When two triangles are similar, then the ratios of the lengths of their corresponding sides are equal.Here, △ABC △DEF $\therefore \frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EF}}$ $\Rightarrow \frac{\mathrm{AB}}{2 \mathrm{AB}}=\frac{6}{\mathrm{EF}}$ $\Rightarrow \mathrm{EF}=12 \mathrm{~cm}$...
Read More →Metal spheres, each of the radius 2 cm,
Question: Metal spheres, each of the radius 2 cm, are packed into a rectangular box of internal dimension 16 cm 8 cm 8 cm when 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid. Solution: The radius of each of the metallic sphere is 2cm. Therefore, the volume of each metallic sphere is $V=\frac{4}{3} \pi \times(2)^{3} \mathrm{~cm}^{3}$ The total volume of the 16 spheres is $V_{1}=16 \times \frac{4}{3} \pi \times(2)^{3} \mathrm{~cm}^{3}$ The internal...
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