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CBSE Questions for Class 11 Medical Chemistry Classification Of Elements And Periodicity In Properties Quiz 11 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Classification Of Elements And Periodicity In Properties
Quiz 11
Which of the following pairs of species has the same electronic configuration?
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$$Zn^{2+}$$ and $$Ni^{2+}$$
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$$Co^{+3}$$ and $$Ni^{4+}$$
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$$Co^{2+}$$ and $$Ni^{2+}$$
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$$Ti^{4+}$$ and $$V^{3+}$$
Explanation
The electron configuration of
$$Zn^{2+}$$is $$1s^22s^22p^63s^23p^63d^{10}$$
$$Ni^{2+}$$ is $$[Ar]3d^8$$
$$Ni^{4+}$$ is $$[Ar]3d^6$$
$$Co^{3+}$$=$$[Ar]3d^6$$
$$Co^{2+}$$= $$[Ar] 4s^23d^7$$
Option B is a correct answer.
Which element has the greatest tendency to lose electrons
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$$F$$
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$$S$$
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$$Fr$$
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$$Be$$
Explanation
Francium belongs to alkaline earth metals having one electron in outermost shell. After losing an electron it will acquire noble gas configuration , so it has the maximum tendency to lose an electron.
Which of the following element has the lowest ionization potential
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$$Fe$$
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$$H$$
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$$Li$$
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$$He$$
Explanation
Li has lowest ionization potential because after losing an electron it will attain noble gas configuration.
The first ionization potential will be maximum for
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Lithium
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Hydrogen
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Uranium
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Iron
Explanation
First ionization energy will be maximum for hydrogen since it has small size. Hydrogen has only one electron and after losing that electron it will be a nucleus with positve charge which is unstable.
Ionization energies in group $$I-A$$ varies in the decreasing order as
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$$Li > Na > K > Cs$$
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$$Na > Li > K > Cs$$
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$$Li > Cs > K > Na$$
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$$K > Cs > Na > Li$$
Explanation
Ionization energy decreses down the group due to increase in atomic size.
So correct order is $$Li > Na > K > Cs$$
The most common oxidation state of an element is $$-2$$. The number of electrons present in its outermost shell is
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4
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2
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6
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8
Explanation
Oxygen has $$6$$ electrons in the outer most shell and shows common oxidation state $$-2$$
The ionization energy of an element is
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The same as the electron affinity of the element
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Equal in magnitude but of opposite sign to the electron affinity of the element
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The energy released when an electron is added to an atoms of the element
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The energy required to remove the outermost electron of an atom of the element
Explanation
Ionization energy is the energy required to remove the outermost electron from gaseous atom or molecule.
In the long from of periodic table, the element having lowest ionisation potentials are present in
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$$I$$ group
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$$IV$$ group
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$$VII$$ group
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Zero group
Explanation
$$I $$ group have the lowest ionisation potential. $$I $$ group belongs to alkali metal and they have lowest ionization potential.
Which of the following gases atom has highest value of $$IE$$
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$$P$$
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$$Si$$
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$$Mg$$
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$$Al$$
Explanation
Across period IE increases as we move from left to right due to decrease in atomic radii. So P has highest value of IE.
Which one of the following elements has the highest ionization energy
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$$[Ne]3s^2 3p^1$$
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$$[Ne]3s^2 3p^2$$
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$$[Ne]3s^2 3p^3$$
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$$[Ar]3d^{10} 4s^2 4p^2$$
Explanation
Option Chas highest ionization energy because it has half-filled p-orbital. Half-filled p-orbital will lead to extra stabilty.
Which among the following species has the highest ionization potential
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$$B$$
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$$Li$$
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$$Ne$$
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$$F$$
Explanation
As we move left to right across a period ionization energy increases.
Ne is a noble gas and has stable electronic configuration. So ionization potential of Ne will be highest.
The electronic configuration of an atom $$A$$ is $$1s^2 2s^2 2p^6 3s^2p^6 3d^{10} 4s^2 4p^3$$. The chemistry of $$A$$ is therefore likely to be similar to that of
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Chlorine
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Nitrogen
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Oxygen
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Boron
Explanation
Nitrogen has similar electronic configuration since it has its valence electrons in $$2s^{2} 2p^{3}$$.
$$1s^2 2s^2 2p^2 3s^2$$ is the electronic configuration of the metal
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$$Na$$
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$$Mg$$
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$$Fe$$
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$$Al$$
Explanation
$$Mg$$ has electronic configuration $$1s^2 2s^2 2p^6 3s^2$$.
The element having the electronic configuration $$1s^2 2s^2 2p^6 3s^2 3p^1$$ is
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A transition element
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A representative element
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An inert gas
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An inner-transition element
Explanation
As last electron is filled in $$3p-$$orbital so it belongs to p-block. s-block and p-block elements are representative elements.
An element has the electronic configuration $$1s^22s^2 2p^6 3s^2 3p^6 3d^5 4s^1$$. It is a
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$$s-$$ block element
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$$p-$$ block element
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$$d-$$ block element
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Inert gas
Explanation
It belongs to d-block as d-orbital is not completley filled. The given electronic configuration belongs to Cr atom and it's a d-block element.
Which one of the following elements has the highest ionization energy
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$$Na$$
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$$Mg$$
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$$C$$
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$$F$$
Explanation
Fluorine has the highest ionization energy because it has tendency to gain an electron rather than losing an electron since it has 7 electrons in outermost shell.
The most important active step in the development of periodic table was taken by
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Mendeleev
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Dalton
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Avogadro
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Cavendish
Explanation
The
most important active step in the development of periodic table was taken
by Mendeleev since though others have also made
significant
contributions towards classification of elements, he elevated the progress made in that direction much more than anyone else.
Hence Correct option is $$A$$.
The element having general electronic configuration $$ 3d^4 4s^1 :$$
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Noble gas
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Non metal
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Metalloid
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Transition metal
Explanation
The element having general electronic configuration $$ 3d^4 4s^1 $$ is a transition metal as its last electron enters d-orbital.
Which of the following can be represented by the configuration $$[Kr]5s^2$$?
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$$Ca$$
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$$Sr$$
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$$Ba$$
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$$Ra$$
Explanation
Electronic Configuration of Strontium is
$$[Kr]5s^2$$.
Hence Option "B" is the correct answer.
The valence shell electronic configuration of alkali metals is
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$$ns^2 np^1$$
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$$ns^1$$
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$$(n-1)p^6ns^2$$
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$$(n-1)d^2ns^2$$
Explanation
The valence shell electronic configuration of alkali metals is
$$ns^1$$
Hence, Option "B" is the correct answer.
The element having electronic configuration belongs to $$ ns^2(n-1)d^{1-10} (n-2)f^{1-14} $$
belongs to :
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s-block
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p-block
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d-block
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f-block
Explanation
The general electronic configuration of f-block is
$$ ns^2(n-1)d^{1-10} (n-2)f^{1-14} $$.
Hence, Option "D" is the correct answer.
The nitride ion in lithium nitride is composed of
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$$7p +7e$$
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$$10p +7e$$
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$$7p +10e$$
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$$10p +10e$$
Explanation
Nitride ion is $$N^{3-}$$. It has $$7$$ protons and $$10$$ electrons
Whose name is not associated with the development of periodic table?
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Prout's
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Newlands
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Rutherford
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Loother Meyer
Explanation
William Prout discovered that the atomic mass of an element can never differ. He observed that many elements have atomic weights as multiple of that of hydrogen.
Newland and Lowther Meyer also played a great role in the attempts to classify the elements.
However, Rutherford played a prominent role in explaining the structure of the atom.
Therefore, the correct option is C.
The alkaline earth metals $$Ba, Sr, Ca$$ and $$Mg$$ may be arranged in the order of their decreasing first ionisation potential as
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$$Mg, Ca, Sr, Ba$$
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$$Ca, Sr, Ba, Mg$$
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$$Sr, Ba, Mg, Ca$$
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$$Ba, Mg, Ca, Sr$$
Explanation
As we go down group electropositive character increases because of increases in size and hence $$I.E.$$ decreases.
$$Ba$$ is the most electropositive element in the group and $$Mg$$ is least.
So, correct order is: $$Mg> Ca> Sr> Ba.$$
The outer electronic configuration of alkaline earth metal is
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$$ns^2$$
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$$ns^1$$
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$$np^6$$
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$$nd^{10}$$
Explanation
The outer electronic configuration of Alkaline Earth metals is "$$\mathrm{ns^2}$$".
Hence, option "A" is the correct answer.
Carbons forms four covalent bonds by sharing its four valence electrons with four univalent atoms , E.g., hydrogen. After the formation of four bonds,carbon attains the electronic configuration of:
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helium
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neon
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argon
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krypton
Explanation
Carbon atom has 2 electrons in its innermost shell and 4 valence electrons. When it forms covalent bonds with 4 univalent atoms, it gets an additional 4 electrons as a result of sharing them. Now there are 8 electrons in its outermost shell. This resembles the electronic configuration of neon, which has the electronic configuration 2, 8.
Option B
The element X has the following electrons configuraton 2, 8, 8, 2 . it means that it belongs to :
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Second period and a second group
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Fourth period and the fourth froup
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Fourth period and the second group
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Second period and fourth group
Explanation
The element which is mentioned is calcium (Ca) it is located in the fourth period and second group.
4th period and 2nd group C is correct answer
Two pairs of electrons are shared between two nitrogen atoms to form a nitrogen molecule
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True
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False
Explanation
Molecular nitrogen consists of two nitrogen atoms triple bonded to each other and, as with all molecules, the sharing of these three pairs of electrons between the two nitrogen atoms allows for the filling of their outer electron shells, making the molecule more stable than the individual nitrogen atoms.
An element having the electronic configuration $$[Ar]3d^24s^2$$ belongs to
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s- block elements
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p- block elements
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d- block elements
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f- block elements
Explanation
An element having the electronic configuration $$[Ar]3d^24s^2$$ belongs to d-block elements as last electron enters in d-orbital.
Consider the following statements:
(A) Electron density in the $$xy-$$ plane in $$3d_{x^2−y^2}$$ orbital is zero.
(B) Electron density in the $$xy-$$ plane in $$3d_{z^2}$$ orbital is zero.
(C) $$2s-$$ orbital has one nodal surface.
(D) For $$2p_x-$$ orbital $$yz$$ is the nodal plane.
Which are the correct statements?
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(C) and (D)
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(B) , and (C)
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Only (B)
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All are correct
Explanation
(A) Electron density in the $$xy-$$ plane in $$3d_{x^2−y^2}$$ orbital is non-zero.
(B) Electron density in the $$xy-$$ plane in $$3d_{z^2}$$ orbital is non-zero.
(C) $$2s-$$ orbital has one nodal surface.
(D) For $$2p_x-$$ orbital $$yz$$ is the nodal plane.
Option C and D are correct.
The electronic configurations of three elements X, Y and Z are X 2, 8; Y 2, 8, 7 and Z 2, 8,Which of the following is correct?
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X is a metal
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Y is a metal
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Z is a non-metal
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Y is a non-metal and Z is a metal
Explanation
Hint: Generally metals have 1 to 3 electrons in the valence shell and non-metals have 4 to 7 electrons in the valence shell and 2 or 8 electrons in the valence shell of inert gas.
Step 1:
Metals have generally 1 to 3 electrons in the valence shell and can easily lose those electrons to form ionic compounds. Generally, non-metals have 4 to 7 electrons in the valence shell and can gain electrons or share electrons to complete their octet. Noble gases have 2 or 8 electrons in the valence shell and are chemically inert.
Step 2:
The electronic configuration of element X is 2, 8. So, it has 8 electrons in the valence shell. It is an inert gas.
The electronic configuration of element Y is 2, 8, 7. So, it has 7 electrons in the valence shell. It is a non-metal.
The electronic configuration of element Z is 2, 8, 2. So, it has 2 electrons in the valence shell. It is a metal.
Step 3:
X is an inert gas since it has a complete octet, Y is a non-metal since it is one electron less to complete octet and Z is metal since it can easily lose 2 electrons in the valence shell.
Final step:
So, the correct option is
(D) Y is a non-metal and Z is a metal.
In Mendeleev's Periodic Table, gaps were left for the elements to be discovered later. Which of the following elements found a place in the periodic table later?
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Germanium
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Chlorine
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Oxygen
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Silicon Ana
Explanation
Mandeleev named the undiscovered elements as Eka-boron, Eka-aluminlum and Eka-silicon. These were replaced with scandium, gallium and germanium respectively in due course of time.
Those elements impart colour to the flame on heating in it, the atoms of which require low energy for the ionisation (i.e., absorb energy in the visible region of spectrum). The elements of which of the following groups will impart colour to the flame?
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$$2$$
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$$13$$
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$$1$$
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$$17$$
Explanation
Group $$I$$ (alkali metals) and Group $$II$$ elements (alkaline earth metals) are reactive metals with low ionization enthalpies.
The electronic configurations of three elements, $$A, B$$ and $$C$$ are given below. What Stable form of $$A$$ may be represented by the formula:
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$$A$$
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$$A_{2}$$
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$$A_{3}$$
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$$A_{4}$$
Explanation
Octate of $$A$$ is complete.
therefore, monoatomic form of $$A$$ is a stable form.
Option A is correct.
The electronic configuration of the outermost shell of the most electronegative element is
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$$2s^{2}2p^{5}$$
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$$3s^{2}3p^{5}$$
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$$4s^{2}4p^{5}$$
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$$5s^{2}5p^{5}$$
Explanation
Elements of Group $$17$$ has $$ns^{2}np^{5}$$ electronic configuration. The electronegativity decreases down the group. The elements of this group are the most electronegative.
The electronic configurations of three elements, $$A, B$$ and $$C$$ are given below. The molecular formula of the compound formed from $$B$$ and $$C$$ will be
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$$BC$$
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$$B_{2}C$$
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$$BC_{2}$$
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$$BC_{3}$$
Explanation
Option D is the correct answer.
$$B$$ represents phosphorus $$P$$ and $$C$$ represents Chlorine $$Cl$$. The compound formed is $$PCl_{3}$$ i.e.,$$BC_{3}$$.
The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds?
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$$ 3 s^{1} $$
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$$ 3 \mathrm{d}^{1} 4 \mathrm{s}^{2} $$
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$$ 3 \mathrm{d}^{2} 4 \mathrm{s}^{2} $$
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$$ 3 s^{2} 3 p^{3} $$
Explanation
Elements that have only s-electrons in the valence shell do not show more than one oxidation state (shows only one oxidation state of +1).
e.g. alkali metals
(b), (c ) having incompletely filled d-orbitals in the outermost shell show variable oxidation states.
Element with outer electronic configuration as $$3d^1\ 4s^2$$ shows variable oxidation states of +2 and +3 and the element with outer electronic configuration as $$3d^2\ 4s^2$$ shows variable oxidation states of +2, +3 and +4.
e.g. transition metals
p-Block elements also show variable oxidation states due to involvement of d-orbitals and inert pair effect.
Thus, element having $$3s^2\ 3p^3$$ as the outer electronic configuration shows variable oxidation states of +3 and +5 due to involvement of d-orbitals.
Hence, option $$B$$, $$C$$, and $$D$$ are correct.
The electronic configurations of three elements, $$A, B$$ and $$C$$ are given below. The bond between $$B$$ and $$C$$ will be
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Ionic
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Covalent
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Hydrogen
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Coordinate
Explanation
Both $$B$$ and $$C$$ are non-metals so, the bond formed between them will be covalent.
The electronic configurations of three elements, $$A, B$$ and $$C$$ are given below. What Stable form of $$C$$ may be represented by the formula:
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$$C$$
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$$C_{2}$$
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$$C_{3}$$
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$$C_{4}$$
Explanation
$$KK[\sigma _{2s}]^{2}[\sigma^*_{2s}]^{2}[\pi _{2px}]^{2}[\pi _{2py}]^{2}$$
Bond order= $$\dfrac{1}{2}[6-2]=2$$
Option B is a correct Answer
$$C_2$$ is diamegnetic in nature.
The electronic configuration of four elements are :
I. $$[Kr]5s^1$$
II.$$[Rn]5f^{14}6d^{1}7s^2 $$
III.$$[Ar]3d^{10}4s^24p^5$$
IV.$$[Ar]3d^64s^2$$.
Consider the following statements :
I show variable oxidation state
II is a d-block element
The compound formed between I and Ill is covalent
IV shows a single oxidation state
Which statement is True (T) or False (F)?
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FTFF
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FTFT
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FFTF
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FFFF
Explanation
(I) $$[Kr]5s^1$$, shows only single oxidation state +1
(II)$$[Rn]5f^{14}6d^{1}7s^2 $$, it is f-block element ($$Z = 103$$)
(III)The compound formed between I and Ill is ionic.
(IV)$$[Ar]3d^64s^2$$, ($$Z = 26$$) Fe shows variable oxidation state.
thus, D is the correct answer.
The outer electronic structure of lawrencium (atomic number $$103$$) is :
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$$[Rn] \ 5f^{13}7s^27p^2$$
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$$[Rn] \ 5f^{13}6d^{1}7s^17p^2$$
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$$[Rn] \ 5f^{14}7s^17p^2$$
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$$[Rn] \ 5f^{14}6d^{1}7s^2$$
Which electronic configuration must represent an atom in an excited state?
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$$1s^2,2s^2,2p^1$$
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$$1s^2,2s^2,2p^2$$
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$$1s^2,2s^2,2p^2,3s^1$$
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$$1s^2,2s^2,2p^5$$
Explanation
$$1s^2,2s^2,2p^2,3s^1$$ (Exited state)
One electrode is excited into S orbital
from P.
Choose the correct option and rewrite the statement.
The number of electrons in the outermost shell of alkali metal is __________.
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$$1$$
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$$2$$
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$$3$$
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$$7$$
Explanation
$$\text{The number of electrons in the outermost shell of alkali metal is 1.}$$
What does 'like dissolves like' mean?
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That compounds that are from the same period on the periodic table will dissolve each other.
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Polar solutes dissolve in polar solvents.
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Nonpolar solutes dissolve in polar solvents.
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Polar solutes dissolve in nonpolar solvents.
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That compounds from the same family on the periodic table will dissolve each other.
Explanation
"Like dissolves like " means if any substance is polar then polar and ionic substances will be dissolved in it. Example : HCl in water. If the solvent is non-polar then the non-polar substances only can be dissolved in it. Example : hexane in benzene.
Electronic configuration of $$Al^{3+}$$ is
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$$2, 8, 3$$
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$$2, 8, 8$$
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$$2, 8$$
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$$2, 8, 8, 3$$
Explanation
The arrangement of electrons in different energy levels around a nucleus is called electronic configuration. The periodicity in properties of elements in any group is due to repetition in the same valence shell electronic configuration after a certain gap of atomic numbers such as 2, 8, 8, 18, 18, 32.
The atomic number of Al is 13 and its electronic configuration is 2,8,3. So, the electronic configuration of Al$$^{3+}$$ is 2,8.
The element with electronic configuration (Ar) $$ 3d^{2} 4s^{2} $$ belongs to which block ?
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s - block
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p - block
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d - block
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f- block
Explanation
The element with electronic configuration $$(Ar) \;3d^24s^2$$ is $$Titanium$$, which belongs to $$d-block$$
hence option (C) is correct.
The outermost electron configuration of an element in this is $$2s^2\ 2p^6$$. This represents:
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Excited state
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Ground state
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Anion form
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Cationic form
The atom of which of the following elements forms dual closed shell configuration by receiving one electron?
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$$He$$
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$$H$$
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$$Li$$
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$$B$$
Explanation
Hydrogen has one electron in its valence shell. It gains one electron and forms a dual closed shell configuration like helium.
Which of the following will not have $$18$$ electrons?
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$$K^+$$
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$$Cl^-$$
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$$K$$
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$$Ca^{2+}$$
Explanation
Atomic number of $$K$$ is $$19$$. Thus, it has $$19$$ electrons.
How many electron are there in chloride ion?
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$$17$$
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$$18$$
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$$16$$
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$$8$$
Explanation
Chloride ion $$(Cl^{-})$$ is formed when chlorine atom $$(17)$$ gains one electron. It thus has $$18$$ electrons.
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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