Extended or long form of periodic table Questions in English

(C) In the modern periodic table,the $13^{th}$ group is designated as group $III$ $A$ (or $13$),not $III$ $B$.
Group $III$ $B$ corresponds to the $3^{rd}$ group in the modern periodic table.
Therefore,the statement in option $C$ is incorrect.

(B) The $3^{rd}$ period of the periodic table contains $8$ elements ($Na$ to $Ar$).
Therefore,the statement that the $3^{rd}$ period contains $18$ elements is incorrect.
The $1^{st}$ period contains $H$ and $He$,both of which are non-metals.
The $7^{th}$ period is incomplete,and while $Og$ $(Oganesson)$ is an inert gas,it is often considered part of the $7^{th}$ period,but historically,the statement refers to the lack of naturally occurring inert gases in that period.
The $p-$block contains all three types of elements: metals,non-metals,and metalloids.

(B) The atomic number of the element is $120$.
According to the periodic table,the noble gas $Og$ (Oganesson) has an atomic number of $118$.
The next element with atomic number $119$ will start the $8^{th}$ period in the $1^{st}$ group $(8s^1)$.
The element with atomic number $120$ will have the electronic configuration $[Og] \, 8s^2$.
Therefore,it belongs to the $2^{nd}$ group and $8^{th}$ period.

(A) The $IUPAC$ nomenclature for elements with atomic number $> 100$ uses specific roots for digits: $0 = nil$,$1 = un$,$2 = bi$.
For atomic number $120$,the roots are $un$ $(1)$,$bi$ $(2)$,and $nil$ $(0)$.
Combining these roots and adding the suffix $-ium$ gives the name $Unbinilium$.
The symbol is derived from the first letters of the roots: $U$ $(un)$,$b$ $(bi)$,and $n$ $(nil)$,resulting in $Ubn$.

$Digit$	$Name-Abbreviation$
$0$	$nil-n$
$1$	$un-u$
$2$	$bi-b$

(N/A) For the $5^{\text{th}}$ period,the principal quantum number is $n=5$.
The orbitals available for filling are $5s$,$4d$,and $5p$.
The number of orbitals in each subshell are: $5s$ ($1$ orbital),$4d$ ($5$ orbitals),and $5p$ ($3$ orbitals).
The total number of orbitals is $1 + 5 + 3 = 9$.
According to the Pauli Exclusion Principle,each orbital can hold a maximum of $2$ electrons.
Therefore,the total number of electrons that can be accommodated is $9 \times 2 = 18$.
Since each element corresponds to the addition of one electron,there are $18$ elements in the $5^{\text{th}}$ period.

(N/A) Following the periodic law and the Aufbau principle,we can predict the positions of these elements:
$1$. For $Z=117$: The element follows the noble gas $Rn$ $(Z=86)$. The configuration is $[Rn] 5f^{14} 6d^{10} 7s^{2} 7p^{5}$. It belongs to Group $17$ (Halogen family).
$2$. For $Z=120$: The element follows the noble gas $Og$ $(Z=118)$. The configuration is $[Og] 8s^{2}$. It belongs to Group $2$ (Alkaline earth metals).

(B) The basic theme of organization of elements in the periodic table is to classify the elements into periods and groups according to their properties.
This arrangement makes the study of elements and their compounds simple and systematic.
In the periodic table,elements with similar properties are placed in the same group.

(17) The $1^{st}$ period contains $2$ elements and the $2^{nd}$ period contains $8$ elements. The $3^{rd}$ period begins after the $2^{nd}$ period,starting with the element having atomic number $Z = 11$. The $3^{rd}$ period contains $8$ elements,ending at $Z = 18$ (which is in the $18^{th}$ group). Therefore,the element in the $17^{th}$ group of the $3^{rd}$ period has an atomic number of $Z = 17$.

The general outer electronic configurations for the elements in different blocks are summarized in the table below:

Element Block	General Outer Electronic Configuration
$s$-block	$ns^{1-2}$
$p$-block	$ns^{2} np^{1-6}$
$d$-block	$(n-1)d^{1-10} ns^{0-2}$
$f$-block	$(n-2)f^{1-14} (n-1)d^{0-1} ns^{2}$

(N/A) $(i)$ Since $n=3$,the element belongs to the $3^{\text{rd}}$ period. It is a $p$-block element since the last electron occupies the $p$-orbital.
There are $4$ electrons in the $p$-orbital. Thus,the group number $= 2 (s\text{-block}) + 10 (d\text{-block}) + 4 (p\text{-electrons}) = 16$.
Therefore,the element belongs to the $3^{\text{rd}}$ period and $16^{\text{th}}$ group. The element is Sulphur $(S)$.
$(ii)$ Since $n=4$,the element belongs to the $4^{\text{th}}$ period. It is a $d$-block element as $d$-orbitals are being filled.
There are $2$ electrons in the $d$-orbital. Thus,the group number $= 2 (s\text{-block}) + 2 (d\text{-electrons}) = 4$.
Therefore,it is a $4^{\text{th}}$ period and $4^{\text{th}}$ group element. The element is Titanium $(Ti)$.
$(iii)$ Since $n=6$,the element is in the $6^{\text{th}}$ period. It is an $f$-block element (Lanthanoid series). All $f$-block elements belong to group $3$.
Its electronic configuration is $[Xe] 4f^{7} 5d^{1} 6s^{2}$. The atomic number is $54 + 7 + 1 + 2 = 64$. The element is Gadolinium $(Gd)$.

(C) The value of the principal quantum number $(n)$ for the outermost shell or the valence shell indicates the period in the Modern periodic table.

(B) The $d$-subshell has $5$ orbitals,which can accommodate a maximum of $10$ electrons $(2 \times 5 = 10)$.
Therefore,the $d$-block consists of $10$ columns.
Statement $(b)$ is incorrect because it states that the $d$-block has $8$ columns.

(N/A) Work of Moseley and Modern Periodic Law:
- When Mendeleev developed his periodic table,chemists knew little about the internal structure of the atom. The beginning of the $20^{th}$ century witnessed profound developments in theories about sub-atomic particles.
- In $1913$,the English physicist Henry Moseley observed regularities in the characteristic $X$-ray spectra of the elements. He established the relation $\sqrt{v} = a(Z - b)$,where $v$ is the frequency of the emitted $X$-ray and $Z$ is the atomic number of the element.
$(i)$ $A$ plot of $\sqrt{v}$ versus $Z$ yields a straight line.
$(ii)$ $A$ plot of $v$ versus $Z$ does not yield a straight line.
- He thereby showed that the atomic number is a more fundamental property of an element than its atomic mass.
- Modern Periodic Law: "The physical and chemical properties of the elements are periodic functions of their atomic numbers."
$(B)$ Modern Periodic Table: The modern periodic table contains $7$ periods and $18$ groups.
- Period: $A$ horizontal row in the periodic table is known as a period.

$Period$	$1, 2, 3, 4, 5, 6, 7$
$Number$ of elements	$2, 8, 8, 18, 18, 32, 32$

- Groups: Vertical columns in the periodic table are known as groups.
- According to the recommendation of the International Union of Pure and Applied Chemistry $(IUPAC)$,the groups are numbered from $1$ to $18$,replacing the older notation of groups ($IA$ to $VIIIA$,$IB$ to $VIIIB$).

(N/A) Atomic number $(Z)$ is defined as the number of protons present in the nucleus of an atom.
The Modern Periodic Law states that the physical and chemical properties of elements are a periodic function of their atomic numbers.
Key aspects of the Modern Periodic Table:
$(i)$ It is based on the electronic configuration of elements,which determines their chemical reactivity and valency.
$(ii)$ It consists of $18$ vertical columns known as groups and $7$ horizontal rows known as periods.
$(iii)$ Elements with similar valence shell electronic configurations are placed in the same group,exhibiting similar chemical properties.

(N/A) Old method of nomenclature and controversy:
$\rightarrow$ The naming of new elements was traditionally the privilege of the discoverer,and the suggested name was ratified by the $IUPAC$. In recent years,this has led to controversy. New elements with very high atomic numbers are so unstable that only minute quantities,sometimes only a few atoms,are obtained.
- Their synthesis and characterization require highly sophisticated,costly equipment and laboratories. Such work is carried out with a competitive spirit in only a few laboratories worldwide.
$\rightarrow$ Scientists,before collecting reliable data on a new element,are sometimes tempted to claim its discovery. For example,both American and Soviet scientists claimed credit for discovering element $104$.
- The Americans named it Rutherfordium,whereas the Soviets named it Kurchatovium.
- To avoid such problems,the $IUPAC$ recommended that until a new element's discovery is proven and its name is officially recognized,a systematic nomenclature be derived directly from the atomic number of the element using numerical roots for $0$ and $1-9$.
- The roots are put together in the order of the digits that make up the atomic number,and "ium" is added at the end.
$(B)$ Digit,Name,and Abbreviation table:

$Digit$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$Name$	nil	un	bi	tri	quad	pent	hex	sept	oct	enn
$Abbreviation$	$n$	$u$	$b$	$t$	$q$	$p$	$h$	$s$	$o$	$e$

(N/A) To name an element with atomic number $106$,we use the $IUPAC$ nomenclature system for elements with $Z > 100$:

Digit	$1, 0, 6$
Root name	$Un, Nil, hex$

Combining these roots and adding the suffix "ium":
$Un + nil + hex + ium = Unnilhexium$.
The symbol is derived from the first letters of the roots: $U + n + h = Unh$.

(N/A) The period number of an element indicates the value of the principal quantum number $(n)$.
As the period number increases,the energy level increases $(n=1, n=2, \ldots \text{ etc. })$.
The number of elements in any period is double the number of atomic orbitals available in that shell.

Principal $(n)$ Energy level	Obtained orbitals	Number of elements
$n=1$ $(K \text{ shell})$	One $(1s)$	$2$ elements: Hydrogen,Helium
$n=2$ $(L \text{ shell})$ Second period	One $2s +$ three $2p$,total four orbitals	$8$ elements: ${}_{3}Li$ to ${}_{10}Ne$
$n=3$ $(M \text{ shell})$ Third period	One $3s +$ three $3p$,total four orbitals	$8$ elements: ${}_{11}Na$ to ${}_{18}Ar$
$n=4$ $(N \text{ shell})$ Fourth period	One $4s +$ five $3d +$ three $4p$,total nine orbitals	$18$ elements: ${}_{19}K$ to ${}_{36}Kr$
$n=5$ $(O \text{ shell})$ Fifth period	One $5s +$ five $4d +$ three $5p$,total nine orbitals	$18$ elements: ${}_{37}Rb$ to ${}_{54}Xe$
$n=6$ $(P \text{ shell})$ Sixth period	One $6s +$ seven $4f +$ five $5d +$ three $6p$,total sixteen orbitals	$32$ elements: ${}_{55}Cs$ to ${}_{86}Rn$ (includes Lanthanide series)
$n=7$ $(Q \text{ shell})$ Seventh period	One $7s +$ seven $5f +$ five $6d +$ three $7p$,total sixteen orbitals	$32$ elements: ${}_{87}Fr$ to ${}_{118}Og$ (includes Actinide series)

(N/A) For the $6^{th}$ period,the principal quantum number $n = 6$.
The orbitals available are determined by the values of $l$ from $0$ to $n-1$. For $n=6$,the subshells involved in filling the period are $6s$,$4f$,$5d$,and $6p$.

Subshell	$6s$	$4f$	$5d$	$6p$
Number of orbitals	$1$	$7$	$5$	$3$

Total number of orbitals $= 1 + 7 + 5 + 3 = 16$.
Since each orbital can hold a maximum of $2$ electrons,the total number of electrons $= 16 \times 2 = 32$.
Therefore,there are $32$ elements in the $6^{th}$ period.

(N/A) Similarity in elements of the same group: Elements in the same group or vertical column have the same valence shell electronic configuration. Elements in the same group have the same number of electrons in their outer orbitals and exhibit similar chemical properties. For example,the valence shell electronic configuration of group-$1$ elements (alkali metals) is $ns^1$.
The names,symbols,and electronic configurations of group-$1$ elements are given in the table below:

Atomic Number and Symbol	Electronic Configuration
$3, Li$	$1s^2 2s^1$ or $[He] 2s^1$
$11, Na$	$1s^2 2s^2 2p^6 3s^1$ or $[Ne] 3s^1$
$19, K$	$1s^2 2s^2 2p^6 3s^2 3p^6 4s^1$ or $[Ar] 4s^1$
$37, Rb$	$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 5s^1$ or $[Kr] 5s^1$
$55, Cs$	$1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6 4d^{10} 5s^2 5p^6 6s^1$ or $[Xe] 6s^1$
$87, Fr$	$[Rn] 7s^1$

(N/A) The element with $Z = 114$ is situated between the noble gases of period $6$ $(Z = 86)$ and period $7$ $(Z = 118)$.
Therefore,its period is $7$.

Order of period	$1, 2, 3, 4, 5, 6, 7$
Last element	$He, Ne, Ar, Kr, Xe, Rn, Og$
$Z$ of last element	$2, 10, 18, 36, 54, 86, 118$

$Z$	$118, 117, 116, 115, 114$
Group	$18, 17, 16, 15, 14$

Electronic Configuration: $[Rn]^{86} \ 5f^{14} \ 6d^{10} \ 7s^{2} \ 7p^{2}$.
The valence shell configuration is $7s^{2} \ 7p^{2}$.
Since the total number of valence electrons is $2 + 2 = 4$,the group number is $10 + 4 = 14$.
Thus,the element is in period $7$ and group $14$.

(N/A)

Period	First Element $(Name/Symbol, Z)$ and Last Element $(Name/Symbol, Z)$
$1$	$Hydrogen (H), 1$; $Helium (He), 2$
$2$	$Lithium (Li), 3$; $Neon (Ne), 10$
$3$	$Sodium (Na), 11$; $Argon (Ar), 18$
$4$	$Potassium (K), 19$; $Krypton (Kr), 36$
$5$	$Rubidium (Rb), 37$; $Xenon (Xe), 54$
$6$	$Cesium (Cs), 55$; $Radon (Rn), 86$
$7$	$Francium (Fr), 87$; $Oganesson (Og), 118$

(N/A) There are four blocks present in the periodic table,which are $s$,$p$,$d$,and $f$ blocks.
This classification depends upon the atomic orbital into which the last electron enters in the ground state of the atom.
$1$. $s$-block: The last electron enters the $s$-orbital.
$2$. $p$-block: The last electron enters the $p$-orbital.
$3$. $d$-block: The last electron enters the $d$-orbital.
$4$. $f$-block: The last electron enters the $f$-orbital.

(A) The elements $Hydrogen$ $(H)$ and $Helium$ $(He)$ are considered exceptions in the periodic table.
Position of $Hydrogen$:
$\rightarrow$ $Hydrogen$ has $1 \ e^-$ in the $1s$ orbital,suggesting it should be in group-$1$.
$\rightarrow$ $Hydrogen$ can accept $1 \ e^-$ to achieve a stable configuration like $Helium$,suggesting it should be in group-$17$.
$\rightarrow$ Due to these conflicting properties,$Hydrogen$ is placed separately in the periodic table.
Position of $Helium$:
$\rightarrow$ $Helium$ has a $1s^2$ configuration,which suggests it should be in the $s$-block.
$\rightarrow$ However,it is placed in the $p$-block with group-$18$ elements because it has a completely filled valence shell,exhibiting properties characteristic of noble gases.

(C) The position of an element in the periodic table is determined by its electronic configuration.
$H$ (Hydrogen) has an atomic number of $1$ and an electronic configuration of $1s^1$. Due to its ability to lose one electron like alkali metals or gain one electron like halogens,it is often placed at the top of group $1$.
$He$ (Helium) has an atomic number of $2$ and an electronic configuration of $1s^2$. Since its valence shell is completely filled (duplet),it exhibits properties of noble gases and is placed in group $18$.

(C) The $p-$block elements consist of groups $13$ to $18$ of the periodic table,where the last electron enters the outermost $p-$orbital.

(N/A) The modern periodic law states that the physical and chemical properties of elements are a periodic function of their atomic numbers. This means that elements with similar electronic configurations repeat at regular intervals,leading to periodicity in their properties.
$1$. Cause of Periodicity: The primary cause is the repetition of similar outer electronic configurations after certain regular intervals.
$2$. Example $1$: Group $1$ (Alkali Metals). All elements in this group have a similar outer electronic configuration of $ns^1$.
- ${}_{3}Li: [He] 2s^1$
- ${}_{11}Na: [Ne] 3s^1$
- ${}_{19}K: [Ar] 4s^1$
- ${}_{37}Rb: [Kr] 5s^1$
- ${}_{55}Cs: [Xe] 6s^1$
- ${}_{87}Fr: [Rn] 7s^1$
Due to this,they all exhibit similar properties,such as being soft,highly reactive,forming basic oxides,and losing one electron to form unipositive ions $(M^+)$.
$3$. Example $2$: Group $17$ (Halogens). All elements in this group have a similar outer electronic configuration of $ns^2 np^5$.
- ${}_{9}F: [He] 2s^2 2p^5$
- ${}_{17}Cl: [Ne] 3s^2 3p^5$
- ${}_{35}Br: [Ar] 3d^{10} 4s^2 4p^5$
- ${}_{53}I: [Kr] 4d^{10} 5s^2 5p^5$
- ${}_{85}At: [Xe] 4f^{14} 5d^{10} 6s^2 6p^5$
Because of this configuration,they all show similar chemical behavior,such as high electronegativity and the tendency to gain one electron to form uninegative ions $(X^-)$.

(N/A) The long form of the periodic table is superior to Mendeleev's periodic table because it classifies elements based on the electronic configuration of atoms rather than atomic mass.
Key characteristics include:
$(i)$ The table consists of $18$ vertical columns (groups) and $7$ horizontal rows (periods).
$(ii)$ It eliminates the anomalies of Mendeleev's table,such as the position of isotopes and the placement of elements with higher atomic mass before those with lower atomic mass.
$(iii)$ Elements are categorized into $s, p, d,$ and $f$-blocks based on the orbital being filled,which provides a clear relationship between electronic structure and chemical properties.
$(iv)$ It provides a systematic arrangement of transition elements (d-block) and inner transition elements (lanthanides and actinides),which were poorly organized in Mendeleev's table.
$(v)$ Periodic trends in properties like atomic radius,ionization enthalpy,and electronegativity are more clearly explained through the modern periodic law.

(N/A) The main basis for the arrangement of elements in the modern periodic table is the electronic configuration of the elements.
According to the modern periodic law,the physical and chemical properties of elements are a periodic function of their atomic numbers $(Z)$.
This means that elements are arranged in increasing order of their atomic numbers,which reflects their electronic configuration,leading to the repetition of similar properties at regular intervals.

(A) In the periodic table,the horizontal rows are called $Periods$. The period number corresponds to the highest principal quantum number $(n)$ of the elements in that row.
The vertical columns are called $Groups$. Elements in the same group have the same valence shell electronic configuration,which leads to similar chemical properties.

(N/A) The number of elements in each period is determined by the filling of orbitals $(ns, (n-2)f, (n-1)d, np)$. The distribution is as follows:

Period	Number of Elements
$1$	$2$
$2$	$8$
$3$	$8$
$4$	$18$
$5$	$18$
$6$	$32$
$7$	$32$ (Incomplete)

(A) The names of elements with atomic numbers greater than $100$ are derived by using numerical roots for each digit of the atomic number in order,followed by the suffix $-ium$.

(C) The maximum number of electrons that can be accommodated in a shell is given by $2n^2$.
Since each element corresponds to the filling of an orbital,the number of elements in a period is equal to the number of electrons that can be accommodated in the shell being filled.
For the $6^{th}$ period,the orbitals being filled are $6s, 4f, 5d,$ and $6p$.
The total number of orbitals is $1 + 7 + 5 + 3 = 16$.
Since each orbital holds $2$ electrons,the maximum number of elements in the $6^{th}$ period is $16 \times 2 = 32$.

(A) In the first period,the principal quantum number is $n = 1$.
For $n = 1$,the azimuthal quantum number $l = (n - 1) = 0$,which corresponds to the $1s$ orbital.
Since an orbital can hold a maximum of $2$ electrons,the first period contains $2$ elements ($H$ and $He$).

(C) The $7^{th}$ period of the periodic table contains the actinoids,which include many man-made (synthetic) radioactive elements.

(B) The elements belonging to the $4f$ (lanthanoids) and $5f$ (actinoids) series are placed outside at the bottom of the extended periodic table. These are known as inner transition elements.

(N/A) The starting elements of each period in the periodic table correspond to the alkali metals (except for period $1$,which starts with Hydrogen). The elements are as follows:

Period	Starting Element
$1$	$H$
$2$	$Li$
$3$	$Na$
$4$	$K$
$5$	$Rb$
$6$	$Cs$
$7$	$Fr$

(C) The modern extended periodic table is divided into $4$ blocks based on the orbital in which the last electron enters.
These blocks are named as $s, p, d$ and $f$ blocks.

(A) Helium $(He)$ is placed in Group $18$ and Period $1$ of the periodic table.
It is placed in Group $18$ because it is a noble gas with a stable valence shell configuration $(1s^2)$.
Although its valence shell is the $K$-shell $(n=1)$,which is full,it shares the characteristic of chemical inertness with other noble gases in Group $18$.

(B) The general outer electronic configurations are as follows:
$s-$ block elements: $ns^{1-2}$
$p-$ block elements: $ns^2 np^{1-6}$
$d-$ block elements: $(n-1)d^{1-10} ns^{0-2}$
$f-$ block elements: $(n-2)f^{1-14} (n-1)d^{0-1} ns^2$

(N/A) The period of an element corresponds to the principal quantum number $(n)$ of the valence shell.
The block of an element is determined by the orbital into which the last electron enters.
The group number is determined by the number of valence electrons for $s$-block and $p$-block elements,or the sum of $(n-1)d$ and $ns$ electrons for $d$-block elements,where $n$ is the principal quantum number of the valence shell.

(N/A) $(i)$ For $n=3$,the element belongs to the $3^{rd}$ period.
Total valence electrons = $2+4 = 6$. Therefore,it belongs to group $16$ ($p$-block element).
$(ii)$ For $n=4$,the element belongs to the $4^{th}$ period.
The configuration is $3d^2 4s^2$. Total valence electrons = $2+2 = 4$. Therefore,it belongs to group $4$ ($d$-block element).
$(iii)$ For $n=6$,the element belongs to the $6^{th}$ period.
The configuration is $4f^7 5d^1 6s^2$. Since the last electron enters the $f$-orbital,it is an $f$-block element (Lanthanoid series).
All $f$-block elements belong to group $3$.

(A) Given the electronic configuration $ns^2 np^4$ with $n = 3$,the configuration is $3s^2 3p^4$.
Since the principal quantum number $n = 3$,the element belongs to the $3^{rd}$ period.
The number of valence electrons is $2 + 4 = 6$.
For $p$-block elements,the group number is calculated as $10 + \text{valence electrons} = 10 + 6 = 16$.
Thus,the element is in period $3$ and group $16$.

(A) The $IUPAC$ nomenclature for elements with atomic number $Z > 100$ is based on the digits of the atomic number: $0=nil, 1=un, 2=bi, 3=tri, 5=pent, 9=enn$.
$(1)$ $Unnilbium$ $(102)$: $Un(1) + nil(0) + bi(2) = 102$ ($No$,$Nobelium$).
$(2)$ $Unnilennium$ $(109)$: $Un(1) + nil(0) + enn(9) = 109$ ($Mt$,$Meitnerium$).
$(3)$ $Unnilpentium$ $(105)$: $Un(1) + nil(0) + pent(5) = 105$ ($Db$,$Dubnium$).
$(4)$ $Unniltrium$ $(103)$: $Un(1) + nil(0) + tri(3) = 103$ ($Lr$,$Lawrencium$).
Therefore,the correct matching is $(1-E, 2-C, 3-A, 4-B)$.

(C) The atomic numbers correspond to the following elements:
$(1)$ Atomic number $101$ is Mendelevium $(Md)$,which matches $(C)$.
$(2)$ Atomic number $104$ is Rutherfordium $(Rf)$,which matches $(A)$.
$(3)$ Atomic number $116$ is Livermorium $(Lv)$,which matches $(E)$.
$(4)$ Atomic number $109$ is Meitnerium $(Mt)$,which matches $(F)$.
Therefore,the correct matching is $(1-C, 2-A, 3-E, 4-F)$.

(D) The $IUPAC$ nomenclature for elements with atomic number $> 100$ uses numerical roots:
$0 = nil$,$1 = un$,$2 = bi$,$3 = tri$,$4 = quad$,$5 = pent$,$6 = hex$,$7 = sept$,$8 = oct$,$9 = enn$.
For unnilennium:
$un = 1$,$nil = 0$,$enn = 9$.
Therefore,the atomic number is $109$.

(A) The $IUPAC$ nomenclature for elements with atomic number $Z > 100$ uses specific roots for each digit: $0 = \text{nil}$,$1 = \text{un}$,$2 = \text{bi}$,$3 = \text{tri}$,$4 = \text{quad}$,$5 = \text{pent}$,$6 = \text{hex}$,$7 = \text{sept}$,$8 = \text{oct}$,$9 = \text{enn}$.
For Unnilunium: $\text{Un} (1) + \text{nil} (0) + \text{un} (1) + \text{ium} = 101$.
Therefore,the atomic number of Unnilunium is $101$.

(A-D) Let us verify the $IUPAC$ names for elements with $Z > 100$:
$1$. Unnilunium $(101)$ is Mendelevium $(Md)$. This is correct.
$2$. Unniltrium $(103)$ is Lawrencium $(Lr)$. This is correct.
$3$. Unnilhexium $(106)$ is Seaborgium $(Sg)$. This is correct.
$4$. Ununnilium $(110)$ is Darmstadtium $(Ds)$. This is correct.
Wait,re-evaluating the systematic nomenclature:
Unnilunium $(101)$ = $Unu$ (Mendelevium).
Unniltrium $(103)$ = $Unt$ (Lawrencium).
Unnilhexium $(106)$ = $Unh$ (Seaborgium).
Ununnilium $(110)$ = $Uun$ (Darmstadtium).
All given matches are actually correct. However,if the question implies an error in the provided options,let us re-check the systematic names.
Actually,all matches provided in the table are correct according to $IUPAC$ nomenclature. Since the question asks to identify the incorrect match and all are correct,there might be a typo in the question source. Given the standard curriculum,all these are correct matches.

(D) $IUPAC$ nomenclature for elements with atomic number $Z > 100$ is based on the digits of the atomic number.
For $119$:
$1$ = un
$1$ = un
$9$ = enn
Suffix = -ium
Therefore,the name is $Ununennium$ $(Uue)$.

(A) The element $E$ is in period $4$ and group $16$. The group $16$ elements are oxygen family (chalcogens).
The element $E$ is Selenium $(Se)$,which has the atomic number $34$ and electronic configuration $[Ar] 3 d^{10} 4 s^{2} 4 p^{4}$.
The element just above $E$ in group $16$ is Sulfur $(S)$,which is in period $3$.
The valence shell electronic configuration of Sulfur $(S)$ is $3 s^{2} 3 p^{4}$.

(D) The atomic number of $Rn$ (Radon) is $86$.
Adding the electrons from the configuration: $86 + 14 (5f) + 1 (6d) + 2 (7s) = 103$.
The element with atomic number $103$ is $Lawrencium$.
According to $IUPAC$ nomenclature for elements with $Z > 100$:
$1 = un$,$0 = nil$,$3 = tri$.
Therefore,the name is $Unniltrium$.