∆ABC is an equilateral triangle of a side 2a units.
Question: ∆ABCis an equilateral triangle of a side 2aunits. Find each of its altitudes. Solution: Let AD, BE and CFbe the altitudes of ∆ABC meeting BC, AC and AB at D, E and F, respectively.Then, D, E and F are the midpoints of BC, AC and AB, respectively.In right-angled∆ABD, we have:AB = 2aandBD =aApplying Pythagoras theorem, we get: $A B^{2}=A D^{2}+B D^{2}$ $A D^{2}=A B^{2}-B D^{2}=(2 a)^{2}-a^{2}$ $A D^{2}=4 a^{2}-a^{2}=3 a^{2}$ $A D=\sqrt{3} a$ units Similarly, $\mathrm{BE}=a \sqrt{3}$ unit...
Read More →How is the variability in oxidation states of transition metals different from that of the non-transition metals?
Question: How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples. Solution: In transition elements, theoxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by 1 (Fe2+and Fe3+; Cu+and Cu2+). In non-transition elements, the oxidation states differ by 2, for example, +2 and +4 or +3 and +5, etc....
Read More →What are interstitial compounds?
Question: What are interstitial compounds? Why are such compounds well known for transition metals? Solution: Transition metalsare large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds....
Read More →What length of a solid cylinder 2 cm in diameter must
Question: What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm? Solution: We are given a solid cylinder of, diameter = 2 cm We have to recast it into a hollow cylinder of length = 16 cm External Diameter = 20 cm and thickness = 2.5 mm=0.25 cm We have to find the height of the solid cylinder that can be used to get a hollow cylinder of the desired dimensions. Volume of a solid cylinder $=\pi r...
Read More →Explain giving reasons:
Question: Explain giving reasons: (i)Transition metals and many of their compounds show paramagnetic behaviour. (ii)The enthalpies of atomisation of the transition metals are high. (iii)The transition metals generally form coloured compounds. (iv)Transition metals and their many compounds act as good catalyst. Solution: (i)Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its s...
Read More →Find the length of altitude AD of an isosceles ∆ABC in which
Question: Find the length of altitudeADof an isosceles ∆ABCin whichAB=AC= 2aunits andBC=aunits. Solution: In isosceles ∆ ABC, we have:AB = AC =2aunits and BC =aunitsLet AD be the altitude drawn from A that meets BC at D.Then, D is the midpoint of BC. $\mathrm{BD}=\mathrm{BC}=\frac{a}{2}$ units Applying Pythagoras theorem in right-angled ∆ABD, we have: $A B^{2}=A D^{2}+B D^{2}$ $A D^{2}=A B^{2}-B D^{2}=(2 a)^{2}-\left(\frac{a}{2}\right)^{2}$ $A D^{2}=4 a^{2}-\frac{a^{2}}{4}=\frac{15 a^{2}}{4}$ $A...
Read More →What are the different oxidation states exhibited by the lanthanoids?
Question: What are the different oxidation states exhibited by the lanthanoids? Solution: In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds....
Read More →In what way is the electronic configuration of
Question: In what way is the electronic configuration of the transition elements different from that of the non-transition elements? Solution: Transition metals have a partially filleddorbital. Therefore, the electronic configuration of transition elements is (n 1)d1-10ns0-2. The non-transition elements either do not have adorbital or have a fully filleddorbital. Therefore, the electronic configuration of non-transition elements is ns1-2or ns2np1-6....
Read More →What are the characteristics of the transition elements and why are they called transition elements?
Question: What are the characteristics of the transition elements and why are they called transition elements? Which of thed-block elements may not be regarded as the transition elements? Solution: Transition elements are those elementsin which the atoms or ions (in stable oxidation state) contain partially filledd-orbital. These elements lie in thed-block and show a transition of properties betweens-block andp-block. Therefore, these are called transition elements. Elements such as Zn, Cd, and ...
Read More →2.2 cubic dm of brass is to be drawn into a cylindrical
Question: 2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire. Solution: The brass volume that has to be drawn into a cylindrical wire is given is $2.2 \mathrm{dm}^{3}=2.2 \times 10^{-3} \mathrm{~m}^{3}$ We have to make a cylindrical wire out of it with diameter =0.25 cm So the radius of this wire $0.125 \times 10^{-2} \mathrm{~m}$ We have to find the length of this wire. Let the length of this wire be We know that the volume of a cylinde...
Read More →What are the characteristics of the transition elements and why are they called transition elements?
Question: What are the characteristics of the transition elements and why are they called transition elements? Which of thed-block elements may not be regarded as the transition elements? Solution: Transition elements are those elementsin which the atoms or ions (in stable oxidation state) contain partially filledd-orbital. These elements lie in thed-block and show a transition of properties betweens-block andp-block. Therefore, these are called transition elements. Elements such as Zn, Cd, and ...
Read More →∆ABC is an isosceles triangle with AB = AC = 13 cm.
Question: ∆ABCis an isosceles triangle withAB=AC= 13 cm. The length of the altitude fromAonBCis 5 cm. FindBC. Solution: It is given that $\triangle A B C$ is an isosceles triangle. Also, AB = AC = 13 cmSuppose the altitude from A on BC meets BC at D. Therefore, D is the midpoint of BC.AD = 5 cm $\triangle A D B$ and $\triangle A D C$ are right-angled triangles. Applying Pythagoras theorem, we have: $A B^{2}=A D^{2}+B D^{2}$ $B D^{2}=A B^{2}-A D^{2}=13^{2}-5^{2}$ $B D^{2}=169-25=144$ $B D=\sqrt{1...
Read More →What is lanthanoid contraction?
Question: What is lanthanoid contraction? What are the consequences of lanthanoidn contraction? Solution: As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because theincrease in nuclear attraction due to the addition of proton is more pronounced than the increase in t...
Read More →What is lanthanoid contraction?
Question: What is lanthanoid contraction? What are the consequences of lanthanoidn contraction? Solution: As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because theincrease in nuclear attraction due to the addition of proton is more pronounced than the increase in t...
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 the two of the balls are 1.5 cm and 2 cm respectively. Determine the diameter of the third ball. Solution: We have one spherical ball of radius 3 cm So, its volume $=\frac{4}{3} \pi(3)^{3}$......(a) It is melted and made into 3 balls. The first ball has radius 1.5 cm So, its volume $=\frac{4}{3} \pi(1.5)^{3}$......(b) The second ball has radius 2 cm So, its volume $=\frac{4}{3} \pi(2)^{3}$.......
Read More →Name the oxometal anions of the first series of the transition metals in which
Question: Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number. Solution: (i) Vanadate, $\mathrm{VO}_{3}^{-}$ Oxidation state of V is + 5. (ii) Chromate, $\mathrm{CrO}_{4}^{2-}$ Oxidation stateof Cr is + 6. (iii) Permanganate, $\mathrm{MnO}_{4}^{-}$ Oxidation state of Mn is + 7....
Read More →What may be the stable oxidation state of the transition element with the following
Question: What may be the stable oxidation state of the transition element with the followingdelectron configurations in the ground state of their atoms : 3d3, 3d5, 3d8and 3d4? Solution:...
Read More →What may be the stable oxidation state of the transition element with the following
Question: What may be the stable oxidation state of the transition element with the followingdelectron configurations in the ground state of their atoms : 3d3, 3d5, 3d8and 3d4? Solution:...
Read More →To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements?
Question: To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples. Solution: The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states (+2 to +7). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in thed-orbital. Howe...
Read More →How many spherical bullets each of 5 cm in diameter
Question: How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11dm 1 m 5 dm? Solution: We are given a metallic block of dimension $=11 \mathrm{dm} \times 1 \mathrm{~m} \times 5 \mathrm{dm}$ We know that, $1 \mathrm{dm}=10^{-1} \mathrm{~m}$ So, the volume of the given metallic block is $=11 \times 10^{-1} \times 1 \times 5 \times 10^{-1}$ $=55 \times 10^{-2} \mathrm{~m}^{3}$ We want to know how many spherical bullets can be formed from this volume of ...
Read More →In the given figure, O is a point inside a ∆PQR such that ∠POR = 90°, OP = 6 cm and OR = 8 cm.
Question: In the given figure,Ois a point inside a ∆PQRsuch that POR= 90,OP= 6 cm andOR= 8 cm. IfPQ= 24 cm andQR= 26 cm, prove that ∆PQRis right-angled. Solution: Applying Pythagoras theorem in right-angled triangle POR, we have: $\mathrm{PR}^{2}=\mathrm{PO}^{2}+\mathrm{OR}^{2}$ $\Rightarrow \mathrm{PR}^{2}=6^{2}+8^{2}=36+64=100$ $\Rightarrow \mathrm{PR}=\sqrt{100}=10 \mathrm{~cm}$ In ∆ PQR, $\mathrm{PQ}^{2}+\mathrm{PR}^{2}=24^{2}+10^{2}=576+100=676$ and $\mathrm{QR}^{2}=26^{2}=676$ $\therefore ...
Read More →Explain briefly how +2 state becomes more and more stable in the first half
Question: Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? Solution: Theoxidation states displayed by the first half of the first row of transition metals are given in the table below. It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, onmoving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also i...
Read More →Explain briefly how +2 state becomes more and more stable in the first half
Question: Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number? Solution: Theoxidation states displayed by the first half of the first row of transition metals are given in the table below. It can be easily observed that except Sc, all others metals display +2 oxidation state. Also, onmoving from Sc to Mn, the atomic number increases from 21 to 25. This means the number of electrons in the 3d-orbital also i...
Read More →A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.
Question: A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut? Solution: Let AB be a guy wire attached to a pole BC of height 18 m. Now, to keep the wire taut let it to be fixed at A.Now, In right triangle ABCBy using Pythagoras theorem, we have $\mathrm{AB}^{2}=\mathrm{BC}^{2}+\mathrm{CA}^{2}$ $\Rightarrow 24^{2}=18^{2}+\mathrm{CA}^{2}$ $\Rightarro...
Read More →How many balls, each of radius I cm,
Question: How many balls, each of radius I cm, can be made from a solid sphere of lead of radius 8 cm? Solution: We are given a solid sphere with radius $R=8 \mathrm{~cm}$. From this sphere we have to make spherical balls of radius $r=1 \mathrm{~cm}$. Let the no. of balls that can be formed be $n$. We know, Volume of a sphere $=\frac{4}{3} \pi r^{3}$. So, volume of the bigger solid sphere $=\frac{4}{3} \pi(8)^{3}$ Volume of one smaller spherical ball $=\frac{4}{3} \pi(1)^{3}$ We know, the volume...
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