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Structure of Atom Test - 46

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Structure of Atom Test - 46
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  • Question 1
    1 / -0
    Study the following table.
    Compound (mol. mass)Compound Mass of the (in grams) taken
    I.$$CO_2(44)$$4.4
    II.$$NO_2(46)$$2.3
    III.$$H_2O_2(34)$$6.8
    IV.$$SO_2(64)$$1.6

    Which two compounds have least mass of oxygen?
    (Molecular masses of compounds are given in brackets)
    Solution
    I. Mass of oxygen present $$=\dfrac{4.4}{44}\times 32 = 3.2\,g$$
    II. Mass of oxygen present $$= \dfrac{2.3}{46}\times 32= 1.6\,g$$
    III.  Mass of oxygen present $$= \dfrac{6.8}{34} \times 32= 6.4\,g$$
     IV. Mass of oxygen present$$= \dfrac{1.6}{64}\times 32 = 0.8\,g$$
    $$\therefore $$ II and IV have least mass of oxygen. 
  • Question 2
    1 / -0
    What is ratio of time periods $$\left( { { T }_{ 1 } }/{ { T }_{ 2 } } \right) $$ in second orbit of hydrogen atom to third orbit of $${He}^{+}$$ ion?
    Solution
    we know that
    time period$$\propto \cfrac { { n }^{ 3 } }{ { t }^{ 2 } }$$
    n=energy level
    t=z=valan enly
    $$\cfrac { { T }_{ H } }{ { T }_{ He } } =\cfrac { { n }_{ 1 }^{ 3 }{ t }_{ 2 }^{ 2 } }{ { n }_{ 2 }^{ 3 }{ t }_{ 1 }^{ 2 } } $$ 
    =$$\cfrac { \left( 8 \right) \times \left( 4 \right)  }{ \left( 27 \right) \times \left( 1 \right)  } $$
    ratio=$$\cfrac { 32 }{ 17 } $$
  • Question 3
    1 / -0
    $$0.25 g$$ of an element $$M$$ reacts with excess fluorine to produce $$0.547 g$$ of the hexafluoride $$MF_6$$. What is the element?
    Solution

    We have

    $$M + F_2 \rightarrow MF_6$$

    moles of M = moles of $$MF_6$$

    let mass of element M = $$x g$$

    therefore we have

    $$\dfrac{0.25}{x} = \dfrac{0.547}{x+114}$$

    so, solving we get , $$x = 95.9 g$$

    therefore, the elemnt is $$Mo$$

  • Question 4
    1 / -0
    Which of the following pairs is correctly matched ? 
    Solution
    The high speed electrons go into the interior of the atoms of the target material and are attracted by the positive charge on their Nuclei. As an electron passes lose to the positive nucleus of an atom in the target the electron is deflected from its path as shown in fig. This result in deceleration of the electron.
    The loss in energy of the electron during deceleration is emitted in the form of X-rays.
  • Question 5
    1 / -0
    Which of the following sets of quanta numbers could represent the last electron added to complete the electronic configuration for a ground state atom of Br (Z = 35)? 
    Solution
    The valence node shell configuration is $$4s^2\,\, 4p^5$$ for the last electron $$n=4, l=1$$.

    $$m=-1, 0$$ or $$ +1, s=+\cfrac {1}{2}$$ or $$-\cfrac {1}{2}$$.

    Hence, $$ 4, 1, 1, -\cfrac {1}{2}$$

    Therefore, option B is correct.
  • Question 6
    1 / -0
    Arrangement of the following group of orbitals in which they fill with electrons :
    (5f, 6p, 4p, 6s, 4d, 4f )
    Solution

    Electrons will first  fill orbitals by the sum of the quantum numbers n and l. Orbitals with equal values of (n+l) will fill with the lower n values first.

    So  energies of $$4p, 4d, 6s, 4f, 6p, 5f$$ are 5,6,6,7,7,8 respectively.
    Hence option A is correct.

  • Question 7
    1 / -0
    A compound has the molecular formula $$X_4O_6$$. If $$10$$ g of $$X_4O_6$$ has $$5.72$$ g X, then the atomic mass of X is :
    Solution
    Molecular weight of $$X_4O_6=(4a+96)$$

    $$\because (4a+96)g$$ $$ X_4O_6$$ contains $$X=4a$$ g

    $$\therefore 10$$g of $$X_4O_6$$ contains X$$=\displaystyle\frac{4a\times 10}{4a+96}$$

    $$5.72=\displaystyle\frac{4a\times 10}{4a+96}$$

    $$22.88a+549.12=40a$$

    $$17.12a=549.12$$

    $$a=32.07$$amu.
  • Question 8
    1 / -0
    According to the Bohr's atomic model, the relation between principal quantum number (n) and radius of orbit(r) is?
    Solution
    Electron angular momentum about the nucleus is an integer multiple of $$\displaystyle\frac{h}{2\pi}$$, where h is Planck's constant.
    Thus, $$I\omega =mvr=\displaystyle \frac{nh}{2\pi}$$
    Hence $$r\propto n$$.
  • Question 9
    1 / -0
    What does the electronic configuration $$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 },3{ s }^{ 1 }$$ indicate?
    Solution
    The atomic number of Neon is $$10$$
    G.S $$Ne\left[ 10 \right] :1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 }\quad $$
    E.S $$Ne\left[ 10 \right] :1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 },3{ s }^{ 1 }\quad $$
    Hence $$1{ s }^{ 2 },2{ s }^{ 2 },2{ p }^{ 5 },3{ s }^{ 1 }$$  electronic configuration indicates the excited state of neon
  • Question 10
    1 / -0
    The ratio of kinetic energy and the total energy of the electron in the nth quantum state of Bohr's atomic model of hydrogen atom is 
    Solution
    The kinetic energy of the electron in the nth state
    $$K=\dfrac {mZ^2e^4}{8\varepsilon ^2_0h^2n^2}$$
    The total energy of the electron in the nth state
    $$T=\dfrac {-mZ^2e^4}{8\varepsilon ^2_0h^2n^2}$$
    $$\therefore \dfrac {K}{T}=-1$$
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