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The d- and f- Block Elements Test - 15

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The d- and f- Block Elements Test - 15
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  • Question 1
    1 / -0
    A complex [MF6]x{\left[ {M{F_6}} \right]^{ - x}} has magnetic moment of 4.91 B.M. whereas [M(CN)6]x{\left[ {M{{(CN)}_6}} \right]^{ - x}} has zero magnetic moment, metal ion can be :
    Solution
    Accordingtoquestion............       fromdatacan  beoncludedthat  firstcomplezcomtains4unpairedelectronasitsmagneticmomentis 4.91BM  magneticmoment(μ)=n(n+2),where  nisno.ofunpairedelectron:Inhighspincomplex,metal   ionhas4unpairedelectronand  inlowspincomplexit  haszerounpairedelectron....so,    Ion  isFe+2,d6t2g4eg2  ishighspin,4unpaired  electron.Fe+2,d6t2g4eg2  lowspin,zerounpairedelectron.            ionis   Fe+2.sothe  correctption  isC.\begin{array}{l} According\, to\, question............ \\ \, \, \, \, \, \, \, from\, data\, can\, \, be\, oncluded\, that\, \, first\, complez\, comtains\, 4\, unpaired\, electron\, as\, its\, magnetic\, moment\, is\,  \\ \, 4.91\, BM\, \, magnetic\, moment\, (\, \mu )=\sqrt { n } (n+2),where\, \, n\, is\, no.of\, unpaired\, electron: \\ In\, high\, spin\, complex\, ,\, metal\, \, \, ion\, has\, 4\, unpaired\, electron\, and\, \, in\, low\, spin\, complex\, it\, \, has\, zero\, unpaired\, electron.... \\ so, \\ \, \, \, \, Ion\, \, is\, F{ e^{ +2 } },{ d^{ 6 } }t_{ 2g }^{ 4 }e_{ g }^{ 2 }\, \, is\, high\, spin,\, 4\, unpaired\, \, electron. \\ F{ e^{ +2 } },{ d^{ 6 } }t_{ 2g }^{ 4 }e_{ g }^{ 2 }\, \, low\, spin\, ,zero\, unpaired\, electron. \\ \, \, \, \, \, \, \, \, \, \therefore \, \, \, ion\, is\, \, \, F{ e^{ +2 } }. \\ so\, the\, \, correct\, ption\, \, is\, C. \end{array}
  • Question 2
    1 / -0
    The ions from among the following which are colourless are 
    (i) Ti4+Ti^{4+}   (ii) Cu+Cu^{+}   (iii) Co3+Co^{3+}    (iv) Fe2+Fe^{2+}
    Solution
    Ti4+[Ar]3d04s0Ti^{4+} \Rightarrow [Ar] 3d^04s^0, No unpaired ee^-
    Cu+[Ar]3d104s0Cu^{+} \Rightarrow [Ar] 3d^{10}4s^0, No unpaired ee^-
    Co3+[Ar]3d64s0Co^{3+} \Rightarrow [Ar] 3d^64s^0, 4 unpaired ee^-
    Fe2+[Ar]3d64s0Fe^{2+} \Rightarrow [Ar] 3d^64s^0, 4 unpaired ee^-

    Ti4+Ti^{4+} and  Cu+Cu^{+} are colourless.
    Option A is correct.
  • Question 3
    1 / -0
    Actinoids comprises of elements from:
    Solution

    Actinium series starts with Actinium (Ac) and ends with Lawrencium(Lr).

    Hence, the correct option is D\text{D}

  • Question 4
    1 / -0
    The chemical similarity between boron and silicon mainly due to equal value of their:
    Solution
    Actually, it is because of diagonal relationship. Such relationship occurs because crossing and descending the periodic table have opposite effects. It is due to ion polarising power mainly. This is defined as charge divided by square of the radius.
  • Question 5
    1 / -0
    Beryllium resembles Aluminium in properties. This is mainly due to:
    Solution
    Beryllium resembles Aluminium in properties. This is due to similar electronegativity, similar atomic size and they diagonal relationship.
  • Question 6
    1 / -0
    If the IP of hydrogen in its ground state is 2.18 x101810^{-18} J/atom, then the electron affinity of Li3+Li^{3+} ion is :
    Solution
    E.A of Li+3=Li^{+3} = -I.P of Li+2Li^{+2}
    I.P of Li+2 =Li^{+2}  = I.P of HZ2n2H \dfrac{Z^2}{n^2}
    Therefore, I.P of Li+2Li^{+2}=2.18×1018J/atom×92.18 \times 10^{-18} J/atom \times 9 as Z=3
    Thus, E.A of Li+3=1.962×1017Li^{+3} =-1.962 \times 10^{-17} J/atom
  • Question 7
    1 / -0
    If the seventh period is completed, the atomic number of the last element would be:
    Solution
    As the seventh period will be completely filled its electronic configuration must be in the manner of noble gas configuration i.e., ns2,np6ns^2,\: np^6 because the last element in every period is a noble gas element. 

    Hence, the electronic configuration of the last element of the seventh period is (Rn) 5f146d107s27p65f^{14} 6d^{10} 7s^27p^6 as the electronic configuration contains 118118 electrons. 

    The last element in the seventh period will have the atomic number 118118.
  • Question 8
    1 / -0
    The element with Z=106Z = 106 belongs to :
    Solution
    Z=106Z=106 is seaborgium and its electronic configaration is (Rn)5f146d47s2(Rn) 5f^{14}6d^{4}7s^{2}. as the differenting electron enter into 7s7s, 7 is the principle quantum number which represents the number of period to which the element belongs hence, the element belongs to 7th period.
  • Question 9
    1 / -0
    Most of the radio active elements are present in:
    Solution
    The actinoids are radioactive elements and the earlier members have relatively long half-lives.

    The most common and known element is Uranium, which is used as nuclear fuel when its converted into plutonium, through a nuclear reaction.

    Option B is correct.
  • Question 10
    1 / -0
    The element californium belongs to a family of :
    Solution
    Hint: It belongs to f-block element. 

    Correct answer: AA

    Explanation for correct option :
    In the periodic table, Actinoid element or the actinide element, any of a series of 1515 consecutive chemical elements from actinium to lawrencium (atomic numbers (89103(89-103).
    The atomic number of californium is 9898 
    Let's write an electronic configuration for this 

    its electronic configuration is Rn867s25f10Rn^{86} 7s^2 5f^{10}

    Hence, it is an f-block element and as it is in the 7th period, it is a part of the actinide series.

    Explanation for correct options :
    • Alkali and alkaline earth metals belong to s-block elements hence these both options are incorrect
    • Lanthanide series have elements 5757 to 7171 but californium has atomic number 9898, hence it not belongs to lanthanide series.
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