Compare the chemistry of the actinoids with that of lanthanoids with reference to:
Question: Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i)electronic configuration (ii)oxidation states and (iii)chemical reactivity. Solution: Electronic configuration The general electronic configuration for lanthanoids is [Xe]544f0-145d0-16s2and that for actinoids is [Rn]865f1-146d0-17s2. Unlike 4forbitals, 5forbitals are not deeply buried and participate in bonding to a greater extent. Oxidation states The principal oxidation state of lanthanoids is (+3)...
Read More →A spherical ball of radius 3 cm is melted and recast into three spherical balls.
Question: A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball. Solution: The radius of the big spherical ball is 3cm. Therefore, the volume of the big spherical ball is $V=\frac{4}{3} \pi \times(3)^{3}$ cubic $\mathrm{cm}$ The radii of the 1stand 2ndsmall spherical balls are 1.5 cm and 2 cm respectively. Therefore, the volumes of the 1stand 2ndspherical balls are respectively $V_{1...
Read More →Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state.
Question: Name the members of the lanthanoid series which exhibit +4 oxidation state and those which exhibit +2 oxidation state. Try to correlate this type of behavior with the electronic configurations of these elements. Solution: The lanthanides that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis. Ce after forming Ce4+attains a stable electronic configuration of [Xe]. Tb after forming Tb4+attains a stable electronic config...
Read More →Use Hund’s rule to derive the electronic configuration of
Question: Use Hunds rule to derive the electronic configuration of Ce3+ion and calculate its magnetic moment on the basis of spin-only formula. Solution: Ce : $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{1} 5 d^{1} 6 s^{2}$ Magnetic moment can be calculated as: $\mu=\sqrt{n(n+2)}$ Where, n= number of unpaired electrons The electronic configuration of Ce3+: 1s22s22p63s23p63d104s24p64d105s25p64f1 In Ce3+,n= 1 $\therefore \mu=\sqrt{1(1+2)}=\sqrt{3}...
Read More →Which is the last element in the series of the actinoids?
Question: Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element. Solution: The last element in the actinoid series is lawrencium, $\mathrm{Lr}$. Its atomic number is 103 and its electronic configuration is $[\mathrm{Rn}] 5 f^{14} 6 d^{1} 7 s^{2}$. The most common oxidation state displayed by it is $+3$; because after losing 3 electrons it attains stable $f^{14}$ configuration....
Read More →A cylindrical tub of radius 12 cm contains water to a depth of 20 cm.
Question: A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical from ball of radius 9 cm is dropped into the tub and thus the level of water is raised byhcm. What is the value ofh? Solution: The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball of radius 9cm into the tub, the height of the raised water ishcm. Therefore, the volume of the raised water is $V=\pi \times(12)^{2} \times h$ cubic $\mathrm{cm}$ The volume of the spherical ball is $V_{1}=\...
Read More →The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids.
Question: The chemistry of the actinoid elements is not so smooth as that of the Lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. Solution: Lanthanoids primarily show three oxidation states (+2, +3, +4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between 4f, 5d, and 6sorbitals is quite large. On the other hand, the energy difference between...
Read More →An aeroplane leaves an airport and flies due north at a speed of 1000 km/hr. At the same time,
Question: An aeroplane leaves an airport and flies due north at a speed of $1000 \mathrm{~km} / \mathrm{hr}$. At the same time, another plane leaves the same airport and flies due west at a speed of 1200 $\mathrm{km} / \mathrm{hr}$. How far apart will the two planes be after $1 \frac{1}{2}$ hours? Solution: Let A be the first aeroplane flied due north at a speed of 1000 km/hr and B be the second aeroplane flied due west at a speed of 1200 km/hr Distance covered by plane A in $1 \frac{1}{2}$ hour...
Read More →What are inner transition elements?
Question: What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95, 102, 104. Solution: Inner transition metals are those elements in which the last electron enters thef-orbital. The elements in which the 4fand the 5forbitals are progressively filled are calledf-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102....
Read More →The internal and external diameters of a hollow
Question: The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm2of the surface is 10 paise. Find the total cost to paint the vessel all over. Solution: We are given the following hemi hollow sphere The internal and external radii of the hollow hemispherical vessel are $\frac{21}{2}=10.5 \mathrm{~cm}$ and $\frac{25.2}{2}=12.6 \mathrm{~cm}$ respectively. Therefore, the total surface area of the hollow hemispherical vessel...
Read More →What are alloys?
Question: What are alloys? Name an important alloy which contains some of thelanthanoid metals. Mention its uses. Solution: An alloy is a solid solutionof two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements. An important alloy of lanthanoids isMischmetal. It contains lanthanoids (9495%), iron (5%), and traces of S, C, Si, Ca, and Al...
Read More →Indicate the steps in the preparation of:
Question: Indicate the steps in the preparation of: (i)K2Cr2O7from chromite ore. (ii)KMnO4from pyrolusite ore. Solution: (i)Potassium dichromate $\left(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\right)$ is prepared from chromite ore $\left(\mathrm{FeCr}_{2} \mathrm{O}_{4}\right)$ in the following steps. Step (1):Preparation of sodium chromate $4 \mathrm{FeCr}_{2} \mathrm{O}_{4}+16 \mathrm{NaOH}+7 \mathrm{O}_{2} \longrightarrow 8 \mathrm{Na}_{2} \mathrm{CrO}_{4}+2 \mathrm{Fe}_{2} \mathrm{O}_{3...
Read More →A hollow sphere of internal and external radii 2 cm and 4 cm
Question: A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone. Solution: The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is $V=\frac{4}{3} \pi\left\{(4)^{3}-(2)^{3}\right\}$ $=\frac{4}{3} \times \frac{22}{7} \times 56$ $=\frac{32 \times 22}{3}$ The hollow sphere is melted to produce a right circular cone of bas...
Read More →ABC is an isosceles triangle, right-angled at B.
Question: ABCis an isosceles triangle, right-angled atB. similar trianglesACDandABEare constructed on sidesACandAB. Find the ratio between the areas of ∆ABEand ∆ACD. Solution: We have, $A B C$ as an isosceles triangle, right angled at $B$. Now, $\mathrm{AB}=\mathrm{BC}$ Applying Pythagoras theorem in right-angled triangle ABC, we get: $\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}=2 \mathrm{AB}^{2} \quad(\because \mathrm{AB}=\mathrm{AC}) \quad \ldots$ (i) $\because \triangle \mathrm{ACD} \sim ...
Read More →A cylindrical jar of radius 6 cm contains oil.
Question: A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres? Solution: The radius of the cylindrical jar is 6cm. The volume of the oil of height 2cm contained in the jar is $V=\pi \times(6)^{2} \times 2$ cubic $\mathrm{cm}$ The radius of each small sphere is 1.5cm. Therefore, the volume of each small sphere is $V_{1}=\frac{4}{3} \times \pi \times(1.5)^{3}$ cubi...
Read More →Give examples and suggest reasons for the following features of the transition metal chemistry:
Question: Give examples and suggest reasons for the following features of the transition metal chemistry: (i)The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. (ii)A transition metal exhibits highest oxidation state in oxides and fluorides. (iii)The highest oxidation state is exhibited in oxoanions of a metal. Solution: (i)In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of th...
Read More →500 persons have to dip in a rectangular tank which is 80 m long and 50 m broad.
Question: 500 persons have to dip in a rectangular tank which is $80 \mathrm{~m}$ long and $50 \mathrm{~m}$ broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is $0.04 \mathrm{~m}^{2}$ ? Solution: The average displacement of water by a person is $0.04$ cubic $m$. Hence, the total displacement of water in the rectangular tank by 500 persons is $V=500 \times 0.04=20$ Cubic $m$. The length and width of the rectangular tank are 80m and 50m res...
Read More →Calculate the number of unpaired electrons in the following gaseous ions:
Question: Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+and Ti3+. Which one of these is the most stable in aqueous solution? Solution: $\mathrm{Cr}^{3+}$ is the most stable in aqueous solutions owing to a $t_{2 \mathrm{~g}}^{3}$ configuration....
Read More →In ∆ABC, AB = AC. Side BC is produced to D. Prove that
Question: In ∆ABC,AB=AC. SideBCis produced toD. Prove that(AD2AC2)=BDCD. Solution: Draw $A E \perp B C$ meeting $B C$ at $D$. Applying Pythagoras theorem in right-angled triangleAED, we get: Since, ABC is an isosceles triangle and AE is the altitude and we know that the altitude is also the median of the isosceles triangle.So,BE = CEandDE+CE = DE + BE = BD $A D^{2}=A E^{2}+D E^{2}$ $\Rightarrow A E^{2}=A D^{2}-D E^{2} \ldots(\mathrm{i})$ In $\triangle A C E$, $A C^{2}=A E^{2}+E C^{2}$ $\Rightarr...
Read More →A cylindrical tub of radius 12 cm contains water to a depth of 20 cm.
Question: A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball dropped into the tub and the level of the water is raised by 6.75 cm. Find the radius of the ball. Solution: The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball into the tub, the height of the raised water is 6.75cm. Therefore, the volume of the raised water is $V=\pi \times(12)^{2} \times 6.75 \mathrm{~cm}^{3}$ Let, the radius of the spherical ball isr. Therefore, the volume of...
Read More →Which metal in the first series of transition metals exhibits +1
Question: Which metal in the first series of transition metals exhibits +1 oxidationstate most frequently and why? Solution: In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because Cu ( +1) has an electronic configuration of [Ar] 3d10. The completely filledd-orbital makes it highly stable....
Read More →A hollow sphere of internal and external diameters 4 cm and 8 cm
Question: A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone. Solution: The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is $V=\frac{4}{3} \pi \times\left\{(4)^{3}-(2)^{3}\right\} \mathrm{cm}^{3}$ The hollow spherical shell is melted to recast a cone of base- radius 4cm. Let, the height of the cone ish. Therefore, th...
Read More →What is meant by ‘disproportionation’?
Question: What is meant by disproportionation? Give two examples of disproportionationreaction in aqueous solution. Solution: It is found that sometimes a relatively less stable oxidation state undergoes an oxidationreduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation. Forexample, Cr(V) is oxidized to Cr(VI) and reduced to Cr(III). Mn (VI) is oxidized to Mn (VII) and reduced to Mn (IV)....
Read More →How would you account for the following:
Question: How would you account for the following: (i)Of thed4species, Cr2+is strongly reducing while manganese(III) is strongly oxidising. (ii)Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii)Thed1configuration is very unstable in ions. Solution: (i)Cr2+is strongly reducing in nature. It has ad4configuration. While acting as a reducing agent, it gets oxidized to Cr3+(electronic configuration,d3). Thisd3configuration can be written a...
Read More →The diameters of the internal and external surfaces
Question: The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm. find the height of the cylinder. Solution: The internal and external radii of the hollow spherical shell are 3cm and 5cm respectively. Therefore, the volume of the hollow spherical shell is $V=\frac{4}{3} \pi \times\left\{(5)^{3}-(3)^{3}\right\} \mathrm{cm}^{3}$ The hollow spherical shell is melted to recast...
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