IMO Practice Test — Classification of Elements and Periodicity in Properties
14 Questions • 15 min • Olympiad level
15:00
Question 1 of 14
If three elements form a Döbereiner triad with end masses 7 and 39, the middle mass is about:
18
23
30
46
Explanation: Mean = $(7+39)/2 = 23$ (sodium).
Question 2 of 14
An element X has the 1st, 2nd, 3rd ionisation enthalpies 496, 4562, 6912 kJ/mol. X most likely belongs to group:
1
2
13
17
Explanation: A huge jump after the 1st IE means one valence electron, so group 1.
Question 3 of 14
How many elements lie in period 6 of the periodic table?
18
32
8
28
Explanation: Period 6 fills 6s, 4f, 5d, 6p = 2+14+10+6 = 32 elements.
Question 4 of 14
Which isoelectronic ion is the largest?
$\text{O}^{2-}$
$\text{F}^-$
$\text{Ne}$
$\text{Na}^+$
Explanation: Among 10-electron species, lowest $Z$ ($\text{O}^{2-}$, $Z=8$) is largest.
Question 5 of 14
An element in period 4 has the highest oxidation state +7. It is most likely:
Ca
Mn
Fe
Zn
Explanation: Mn ($3d^54s^2$) reaches +7 (as in $\text{MnO}_4^-$).
Question 6 of 14
Identify the property that increases across a period and decreases down a group:
atomic radius
metallic character
electronegativity
number of shells
Explanation: Electronegativity increases across a period and decreases down a group.
Question 7 of 14
If element Y has valence configuration $3s^23p^4$, its highest oxide formula is:
$\text{YO}$
$\text{YO}_2$
$\text{YO}_3$
$\text{Y}_2\text{O}_7$
Explanation: Y is sulphur (group 16); highest oxide $\text{SO}_3$, i.e. $\text{YO}_3$.
Question 8 of 14
Two elements with $Z=11$ and $Z=17$ combine. The expected formula is:
$\text{XZ}$
$\text{X}_2\text{Z}$
$\text{XZ}_2$
$\text{X}_2\text{Z}_3$
Explanation: Na (+1) and Cl (-1) give NaCl, i.e. a 1:1 formula.
Question 9 of 14
The element whose IUPAC symbol is Unp ($Z=105$) was later named:
rutherfordium
dubnium
seaborgium
bohrium
Explanation: $Z=105$ unnilpentium is dubnium.
Question 10 of 14
Which sequence shows correctly increasing first ionisation enthalpy?
B < Be < C < N
Be < B < C < N
N < C < B < Be
C < N < Be < B
Explanation: Order is B(801) < Be(899) < C(1086) < N(1402) kJ/mol due to the B
Question 11 of 14
An oxide $\text{M}_2\text{O}$ turns red litmus blue. Element M is most likely a:
non-metal
metalloid
group 1 metal
noble gas
Explanation: A basic oxide of formula $\text{M}_2\text{O}$ points to an alkali (group 1) metal.
Question 12 of 14
Element P resembles Mg in many reactions though it is in period 2. P is:
Be
Li
B
C
Explanation: Li shows a diagonal relationship with Mg.
Question 13 of 14
Across period 2 the number of valence electrons changes from Li to Ne as:
1 to 7
1 to 8
2 to 8
1 to 6
Explanation: Li has 1 and Ne has 8 valence electrons.
Question 14 of 14
For the set $\text{S}^{2-}, \text{Cl}^-, \text{K}^+, \text{Ca}^{2+}$ (all 18 e), the smallest is:
$\text{S}^{2-}$
$\text{Cl}^-$
$\text{K}^+$
$\text{Ca}^{2+}$
Explanation: Isoelectronic; highest $Z$ ($\text{Ca}^{2+}$, $Z=20$) is smallest.
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