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7801.

An element X which is a yellow solid at room temperature shows catenation and allotropy. X forms two oxides which are also formed during the thermal decomposition of ferrous sulphate crystals and are the major air pollutants. (a) Identify the element X. (b) Write the electronic configuration of X. (c) Write the balanced chemical equation for the thermal decomposition of ferrous sulphatecrystals. (d) What would be the nature (acidic/basic) of oxides formed? (e) Locate the position of the element in the Modern Periodic Table.

Answer»

SOLUTION :(a) Element X is SULPHUR (atomic no. 16)
`underset(2)(K), underset(8)(L), underset(6)(M)` (d) ACIDIC(e) 3rd PERIOD, group 16
Discussion
7802.

An element X placed in 2nd group and 4th period of periodic table burns in the presence of oxygen to form a basic oxide i) Identify the element ii) Write its electronic configuration iii) Write a balanced equation for the reaction when this oxide is dissolved in water

Answer»

Solution :i] The ELEMENT is calcium
ii] `{:(K, L, M, N),(2,8,8,2):}`
iii]`2 CaO(s)+H_2O(l) to Ca(OH)_2 (aq)`
Discussion
7803.

An element X on exposure to moist air turns reddish brown and new compound Y is formed. Identify X and Y.

Answer»

X = Fe, `Y = Fe_(2)O_(3)`
`X = AG, Y = Ag_(2)S`
X= Cu, Y = CUO
`X = Al, Y= Al_(2)O_(3)`

SOLUTION :X = Fe, `Y = Fe_(2)O_(3)`
Discussion
7804.

An element X of group 15 and period 2 exists as diatomic molecule. It combines with hydrogen and forms ammonia in the presence of a suitable catalyst. Draw the electron dot structure of ammonia.

Answer»

SOLUTION :`H XX * underset(H)underset(xx *)OVERSET(* *)N * xx H`
Discussion
7805.

An element X on reacting with oxygen forms an oxide X_2O. This oxide dissolves in water and turns blue litmus red. State whether element X is metal or non-metal

Answer»

SOLUTION :Non-metal.
Discussion
7806.

An element X of group 15 exists as diatomic molecule and combines with hydrogen at 773 K in presence of the catalyst to form a compound, ammonia which has a characteristic pungent smell (a) Identify the element X. How many valence electrons does it have ? (b) Draw the electron dot structure of the diatomic molecule of X. What type of bond is formed in it ? (c) Draw the electron dot structure for ammonia and what type of bond is formed in it ?

Answer»

Solution : (a) Nitrogen (ATOMIC no. 7)
Electronic configuration: 2, 5, it has 5 VALENCE ELECTRONS.
Discussion
7807.

An element X of group 15 and period 2 exists as diatomic molecule. It combines with hydrogen and forms ammonia in the presence of a suitable catalyst. Draw the electron dot structure of the diatomic molecule, and identify and name the type of bond formed.

Answer»

SOLUTION :`OVERSET(* *)(N)vdot vdot overset(* *)N` Triple COVALENT BOND `(N-=N)`
Discussion
7808.

An element X of group 15 and period 2 exists as diatomic molecule. It combines with hydrogen and forms ammonia in the presence of a suitable catalyst. Identify the element X.

Answer»

SOLUTION :X is NITROGEN
Discussion
7809.

An element X of group 15 and period 2 exists as diatomic molecule. It combines with hydrogen and forms ammonia in the presence of a suitable catalyst. Write the electronic configuration of X and count the number of valence electrons.

Answer»

SOLUTION :X =2,5 , NUMBER of VALENCE electrons =5
Discussion
7810.

An element .X. is forming an acidic oxide. Its position in modern Periodic Table will be

Answer»

Group 1 and Period 3
Group 2 and Period 3
Group 13 and Period 3
Group 16 and Period 3

Solution :Non-metals which are on extreme R.H.S. of the Periodic Table FORM acidic OXIDES.
Discussion
7811.

An element 'X' has three completely filled shells and forms a stable uninegative ion and the total number of electrons in the third and fourth shells is same. Identify the elements that form unipositive and dipositive ions that are isoelectronic with X^(-) ion. Compare the reactivities of these two elements with water. Give a reason in support of your answer.

Answer»

Solution :The element 'X' must be a halogen since ot forms
the stable `X^(-)` ion. As there are three completely
filled shells in 'X', the ELECTRONIC configuration of
'X' could be 2, 8, 18, 18, 7 and `X^(-)` is 2, 8, 18, 18, 8. Since there are 54 `e^(-)s"in" X^(-)`, the CORRESPONDING
elements that form unipositive and dipositive ions
should be cesium (55) and barium (56).Cesium
REACTS more vigorously with water than barium.
This is because reactivity of alkali metals is more
due to their greater atomic size when compared
with alkaline earth metals. Moreover, alkaline earth
metals have greater IP values than alkali metals.
Hence, they react slowly with water.
Discussion
7812.

An element 'X' has mass number 35 and number of neutrons 18. Write atomic number and electronic configuration of 'X'. Also write group number, period number and valency of 'X'.

Answer»

Solution :Atomic number of X = Mass number of X - Number of neutrons = 35 – 18 = 17
Therefore, electronic configuration of X = 2, 8, 7
Group number = 17
Period number = 3
Valency = 8 - 7 = 1
Discussion
7813.

An element 'X' has an atomic number 11 while an element 'Y' has an atomic number 17. They are expected to form

Answer»

an IONIC compound with formula XY
an ionic compound with formula `XY_(2)`
a COVALENT compound with formula `XY_(2)`
a covalent compound with formula `X^(2)Y`

SOLUTION :`""_(11)X""2,8,1`
`""_(17)Y""2,8,7`
The bond formed between X and Y will be ionic.
Discussion
7814.

An element 'X' has atomic number 13. a] Write its electronic configuration. b] State the group to which 'X' belongs. c] Is 'X' a metal or non metal? d] Write the formula of its bromide.

Answer»

Solution :a] Electronic configuration is 2, 8, 3
b] It belongs to group 13
c] .X. is a metal
d]
`XBr_3` is the FORMULA of its BROMIDE.
Discussion
7815.

An element X has mass number 35 and number of neutrons = 18. Write atomic number and electronic configuration of X. Also write group number, period number and valency of X.

Answer»

Solution :Atomic number = mass number - number of neutrons `= 35 - 18 = 17. `
Electronic configuration of `X: 2, 8, 7`
GROUP number = 17, period number = 3, valency = 8 - 7 = 1
Discussion
7816.

An element X forms an oxide X_(2)O_(3). In which group of Mendeleev's periodic table is this element placed ?

Answer»

GROUP II
Group III
Group V
Group VIII
Discussion
7817.

An element X forms an oxide which turns red litmus blue. Identify whether X is a metal or non-metal.

Answer»

SOLUTION :METAL, because metal OXIDES are BASIC in NATURE.
Discussion
7818.

An element 'X' belongs to third period and second group of the Modern Periodic Table. (i) Write its electronic configuration. (ii) Is it a metal or non-metal? Why? (iii) Write the formula of the compound formed when 'X' reacts with an element (a) Y of electronic configuration 2, 6 and (b) Z with electronic configuration 2, 8, 7.

Answer»

Solution :(i) Electronic CONFIGURATION - 2,8,2.
(ii) METAL, as it can EASILY LOSE electrons (from outer most orbit)
(III)
Discussion
7819.

An element 'X' belongs to the 3rd period and group 16 of the Modern Periodic Table. a) Determine the number of valence electrons and the valency of 'X'. b] Write the molecular formula of the compound when X reacts with hydrogen and write its electron-dot structure. c] Name the element X and state whether it is metallic or non-metallic

Answer»

SOLUTION :(i) Electronic CONFIGURATION of X = 2,8,6.
Valence ELECTRONS = 6.
Valency = 8-6 = 2.
(ii) Formula with hydrogen - `H_(2)XorH_(2)S`

(III) Sulphur, non-metal.
Discussion
7820.

An element 'X' (Atomic number = 20) burns in the presence of oxygen to form a basic oxide. (i) Identify the element and write its electronic configuration. (ii) State its group number and period number in the Modern Periodic Table. (iii) Write a balanced chemical equation for the reaction when this oxide is dissolved in water.

Answer»

SOLUTION :(i) X(20) --- 2, 8, 8, 2. It is CALCIUM.
(II) Second group, fourth PERIOD.
(III) `CaO+H_(2)OtoCa(OH)_(2)`
Discussion
7821.

An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a divalent halide. (a) What is the position of elements X and Y in the periodic table ? (b) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed.

Answer»

Solution :Element X is a non-METAL while element Y is a metal.
Molecular formula of dihalide is `YX_(2)`.
(a)
(B) BASIC oxide : YO NATURE of the bond : Ionic
Discussion
7822.

An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a Compound. i]Write the position of these elements in the Modern Periodic Table ii] Which one is a metal and which one is a non-metal? iii] Write the formula of the compound.

Answer»

Solution :Electronic CONFIGURATIONS are: `X (17) to 2, 8, 7 and Y (20) to 2, 8, 8, 2. 1`
i] X belongs to group 17 and period 3. Y belongs to group 2 and period 4.
ii] Y is a metal and X is a non-metal.
III] YX, is the formula of the compound formed by them SINCE valency of X is ONE and Y is two.
Discussion
7823.

An element X (atomic number 17) reacts with an element Y (atomic number 20) to forma divalent halide.(a) Where in the Periodic Table are elements X and Y placed? (b) Classify X and Y as metal (s), non-metal (s) or metalloid (s). (c ) What will be the nature of oxide of element Y? Identify the nature of bonding in the compound formed. (d) Draw the electron dot structure of the divalent halide.

Answer»

Solution : (a) X BELONGS to Group 17 and 3rd PERIOD
Y belongs to Group 2 and 4th period
(b) X-Non-metal and Y-Metal
(C) Basic oxide, Ionic bonding
Discussion
7824.

An element X (atomic number 17) reacts with an element Y (atomic number 20) to form a compound. (i) Write the position of these elements in the modern periodic table. (ii) Write the formula of the compound formed. Justify your answer in each case.

Answer»

Solution :(i) The position of elements X in the modern periodic table is group number 17 and PERIOD number 3. The position of element Y in the modern periodic table is group number 2 and period number 4. Electronic configuration of element X = 2, 8,7.
It has 3 shells so period number 3.
Halogens are kept in group 17.
Electronic configuration of element Y = 2, 8, 8,2
It has 4 shells so period number is 4.
The VALENCE shell has 2 electrons so the group number is 2.
(ii) Formula of the compound :
`{:("Valency of element Y","Valency of element X"),("2","1"):}`
`thereforeYX_(2)`
Discussion
7825.

An element X (2, 8, 2) combines separately with SO_(4)^(2-),PO_(4)^(3-) radicals. (i) Write the formulae of the two compounds so formed. (ii) To which group of the modern periodic table does the element X belong? (iii) Will it form covalent or ionic compound ? Give reasons.

Answer»

Solution :(i) FORMULAE of these compounds are `XSO_(4),X_(3)(PO_(4))_(2)`.
(ii) It belongs to group `2^(nd)AND3^(rd)` period.
(iii) It will form ionic compounds as it can lose TWO electrons to acquire stable electronic configuration.
Discussion
7826.

An element with low ionization energy combines with an element having high electron affinity to form

Answer»

ionic BOND
COVALENT bond
dative bond
NONE of these

ANSWER :a
Discussion
7827.

An element with density 2.8 gcm^(-3) forms a f.c.c unit cell with edge length 4xx10^(-8) cm. calculate the molar mass of the element. Given: (N_(A)=6.022xx10^(23)) mol^(-1)

Answer»

26.97
22
25
21

Answer :A
Discussion
7828.

An element which never has a positive oxidation state in any of its compound is

Answer»

Boron
Oxygen
Chlorine
Florine

Answer :D
Discussion
7829.

An element which is an essential constituent of all organic compounds belongs to …………… group .

Answer»

14TH
15TH
16TH
17th

Answer :A
Discussion
7830.

An element which is an essential constituent of all organic compounds belong to ........

Answer»

GROUP 4
Group10
Group 16
Group 14

Answer :A::D
Discussion
7831.

An element which is an essential constituent of all organic compounds belongs to

Answer»

GROUP 1
Group 14
Group 15
Group 16

Answer :B
Discussion
7832.

An element reacts with oxygen to give a compound with a high melting point. This compoundis also soluble in water. The element is likely to be

Answer»

Calcium
Carbon
SILICON
Iron

Solution : Calcium reacts with oxygen to GIVE CaO which dissolves in water to give lime water.
`CA+ 1/2 O_2 to CaO`
`CaO + H_2O to Ca(OH)_2`
Discussion
7833.

An element reacts with oxygen to give a compound with a high melting point. This compound is also soluble in water. The element is likely to be ...

Answer»

CALCIUM
CARBON
SILICON
iron

Solution :calcium
Discussion
7834.

An element reacts with oxygen to give a compound with a high melting point . This compound is also soluble in water, The element is likely to be:

Answer»

calcium
carbon
silicon
iron

Solution :Pure calcium oxide or lime, is an amorphous white solid, having a HIGH melting point (2273 k)
When lime is added to water, a hissing SOUND is produced and a large amount of heat is GENERATED.
`CaO+H_2O to Ca(OH)_2_(Slaked lime)+Heat`
Therefore, the CORRECT answer is (a)
Discussion
7835.

An element reacts with oxygen to give a compound with high melting point. This compound is also soluble in water. The element is likely to be

Answer»

CALCIUM
Carbon
Silicon
Iron

Solution :Calcium
Discussion
7836.

An element placed in 2nd Group and 3rd Period of the Periodic Table, burns in presence of oxygen to form a basic oxide. (a) Identify the element. (b) Write the electronic configuration. (c) Write a balanced equation when it burns in the presence of air. (d) Write a balanced equation when this oxide is dissolved in water. (e) Draw the electron dot structure for the formation of this oxide.

Answer»

SOLUTION : (a) MAGNESIUM (MG)
Discussion
7837.

An element P (atomic number 20) reacts with an element Q (atomic number 17) to form a compound. Answer the following questions giving reason: Write the position of P and Q in the Modern Periodic Table and the molecular formula of the compound formed when P reacts with Q.

Answer»

Solution :CHANGE X to Q and Y to P in the above answer. Formula of the COMPOUND is `PQ_2`
Discussion
7838.

An element of group 14 has two common allotropes, A and B. A is very hard and is bad conductor of electricity while B is soft to touch and good conductor of electricity. Identify the element and its allotropes. Explain reasons for their different properties.

Answer»

Solution :The element is carbon and the two ALLOTROPES are DIAMOND and graphite. Diamond has three-dimensional rigid structure and does not have any free ELECTRONS. Hence, it is hard and bad conductor of electricity. Graphite forms hexagonal sheet-like structure and ONE valency (one electron) with carbon is free. Hence, graphite is soft and a good conductor of electricity.
Discussion
7839.

An element 'M' with electronic configuration (2,8,2) combines separately with NO_(3)^(-),SO_(4)^(2-)andPO_(4)^(3-) radicals. Write the formula of the three compounds so formed. To which group and period of the modern periodic table does the element 'M' belong ? Will 'M' form covalent or ionic compounds? Give reason to justify your answer.

Answer»

Solution :(i) The electronic configuration (2, 8, 2) of the element .M. suggests that it belongs to GROUP 2 and period 3 of the modern periodic table and its VALENCY is 2.
(ii) The chemical FORMULA of the compounds are : `M(NO_(3))_(2)//Mg(NO_(3))_(2),MSO_(4)//MgSO_(4),M_(3)(PO_(4))_(2)//Mg_(3)(PO_(4))_(2)`.
(iii) .M. will form ionic compounds by LOSING two electrons.
Discussion
7840.

An element 'M' with electronic configuration (2, 8, 2) combines separately with (NO_(3))^(-), (SO_(4))^(2-) and (PO_(4))^(3-) radicals. Write the formula of the three compounds so formed. To which group and period of the Modem Periodic Table does the element 'M' belong ? Will 'M' form covalent or ionic compounds ? Give reason to justify your answer.

Answer»

SOLUTION :The element M has 2 electrons in the last orbit. Thus it FORMS `M^(2+)` or it has a VALENCY of `+2`.

The element M belongs to Group 2 and period 3 of the Modem Periodic Table.
The element M will form IONIC COMPOUNDS. It has a tendency to lose 2 electrons to gain stable octet configuration. Thus it will form ionic bonds and give rise to ionic compounds.
Discussion
7841.

An element M with electronic configuration (2,8, 2) combines separately with (NO_3)^(-1),(SO_4)^(2-) and (PO_4)^(3-) radicals. Write the formula of the three compounds so formed. To which group and period of the Modern Periodic Table does the element Mbelong? Does M form covalent or ionic compounds? Justify your answer.

Answer»

Solution :The element M has electronic configuration (2, 8, 2), it has two VALENCE electrons and a VALENCY of +2.
Hence, it forms COMPOUNDS `M(NO_3)_(2)^(-), MSO_4`, and `M_3 (PO_4)_2`, with radicals `(NO_3), (SO_4)^(2-) and (PO_4)^(3-)` respectively.
SINCE M has two valence electrons, it belongs to the 2nd group, and since it has three shells, it belongs to the 3rd period of the PERIODIC table.
M forms ionic compounds since it is a metal.
Discussion
7842.

An element M has atomic number 12. (i) Write the electronic configuration and valency. (ii) Is M a metal or non-metal ? Give reason in support of your answer. (iii) Write the formula and nature (acidic/basic) of the oxide of M.

Answer»

Solution :(i) Electronic configuration of M
`K=2, L=4`
VALENCY = 4
(ii) M is a non-metal because it cannot LOSE the ELECTRONS easily.
(III) `CO_(2)`. It is ACIDIC in nature.
Oxides of non-metals are acidic in nature.
Discussion
7843.

An element is placed in 2nd group and 3rd period of the periodic table, burn in presence of oxygen to form a basic oxide : (i) Identify the element (ii) Write the electronic configuration (iii) Write the balanced equation when it burns in the presence of air. (iv) Write a balanced equation when this oxide is dissolved in water. (v) Draw the electron dot structure for the formation of this oxide.

Answer»

Solution :(i) Magnesium.
(II) Electronic configuration is 2,8, 2
(iii) `2Mg+O_(2)to2MgO`
(iv) `MgO+H_(2)OtoMg(OH)_(2)`
(V) `MG toMg^(2+)+2e^(-)`
2,8,22, 8
`O+2e^(-)toO^(-2)`
2,62,8
is electron dot STRUCTURE.
Discussion
7844.

An element is placed in 2nd Group and 3rd period of Periodic Table, it burns in the presence of oxygen to form basic oxide. (a) Identify the element. (b) Write the electronic configuration. (c) Write the balanced equation when it burns in the presence of air. (d) Write a balanced equation when this oxide is dissolved in water. (e) Draw the electron dot structure for the formation of this oxide.

Answer»

Solution :(a) The ELEMENT is magnesium `(Mg)`.
(b) Electronic configuration of Mg is `{:(K,L,M),(2,8,2):}`
(c) The balanced equation when it bums in the presence of AIR.
`2Mg+O_(2) to 2Mg O`
(d) `MgO+ H_(2) O to Mg (OH)_(2)`
(E )
Discussion
7845.

An element is placed in 5th period and 3rd group what is its atomic number?

Answer»


ANSWER :39
Discussion
7846.

An element in period 3 whose electron affinity is zero.

Answer»

Neon
Sulphur
SODIUM
ARGON

SOLUTION :Argon
Discussion
7847.

Element with electronic configuration as [Ar]^(18)3d^(5)4s^(2) is placed in :

Answer»


ANSWER :s -
Discussion
7848.

An element having atomic number 3.5 can be placed between Be and B.

Answer»


ANSWER :FALSE
Discussion
7849.

An element has same number of electrons in I and IV shells and in the II and III shells i] Write its electronic configuration ii] Write the group and period it belong iii] What is the valency of the element

Answer»

Solution :i] Electronic configuration `{:(K,L,M,N),(2,8,8,2):}`
ii] GROUP 2 [because valence SHELL contains 2 ELECTRONS] PERIOD 4th (as the number of shells are 4]
iii] Valency of the element is 2.
Discussion
7850.

An element has its electronic configuration as 2, 8, 2. Now anwswer the following questions : With which of the following elements would this element resemble? (Atomic numbers are given in the brackets) N (7), Be (4), Ar (18), Cl (17)

Answer»

Solution :The atomic number of the GIVEN element is 12 and thi s element is Magnesium which belongs togroup 2 and Period 3.
E l ectronic configuration of Mg (12) = (2, 8, 2)
Electronic configuration of N (7)= (2, 5)
Electronic configuration of Be (4)= (2, 2)
Electronic configuration of Ar (18) = (2, 8, 8)
Electronic configuration of Cl (17) = (2, 8, 7)
Number of va lence electrons in this element (Magnesium) is 2. Similarly the number of valence e lectron in Be is 2. As they have same number of valence electrons they belong to same group i.e. group 2 and will resemble in some of their PROPERTIES.
Discussion